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Calculus I  ·  Chapter 6  ·  Section 6.1

Integration by Parts
& Integral Tables

A powerful technique for products of functions

When substitution fails, integration by parts steps in — transforming a hard integral into a manageable one using the product rule in reverse.

Why This Section Matters

Not Every Integral Yields
to Substitution

You have already mastered substitution — the technique that "un-does" the chain rule. But what about integrals like $\int x e^x\,dx$ or $\int x \ln x\,dx$? These are products of two different types of functions, and no substitution will untangle them.

Integration by parts is the answer. It is used everywhere in applications: computing present values of income streams, solving differential equations that model population growth, and evaluating probabilities in statistics. Master this technique and a whole new class of real-world problems opens up.

Real uses you will see in this course: Computing the present value $\int_0^T f(t)e^{-rt}\,dt$ when $f(t)$ is not constant — this requires integration by parts. Also: solving logistic differential equations, finding areas under curves like $y = x^2 e^{-x}$, and evaluating integrals that appear in business growth models.
§ 1 — Where the Formula Comes From

Integration by Parts:
The Product Rule in Reverse

You know the product rule for differentiation. Integration by parts is what you get when you integrate both sides of it.

Start with the product rule. If $u$ and $v$ are both functions of $x$:

$$\frac{d}{dx}[uv] = u\frac{dv}{dx} + v\frac{du}{dx}$$

Integrate both sides with respect to $x$:

$$\int \frac{d}{dx}[uv]\,dx = \int u\frac{dv}{dx}\,dx + \int v\frac{du}{dx}\,dx$$

The left side simplifies (integration undoes differentiation):

$$uv = \int u\,dv + \int v\,du$$

Rearrange to solve for $\int u\,dv$:

$$\boxed{\int u\,dv = uv - \int v\,du}$$

This is the integration by parts formula. The idea: if $\int u\,dv$ is hard, transform it into $uv - \int v\,du$ and hope that $\int v\,du$ is easier. ∎

Integration by Parts Formula

If $u$ and $v$ are differentiable functions of $x$, then:

$$\int u\,dv = uv - \int v\,du$$

Equivalently: $\int f(x)g'(x)\,dx = f(x)g(x) - \int g(x)f'(x)\,dx$

§ 2 — The Step-by-Step Procedure

How to Use
Integration by Parts

Step 1
Choose u and dv

Split the integrand into two parts: one you call u, one you call dv. The choice matters — u should become simpler when differentiated, and dv should be easy to integrate.

Step 2
Compute du and v

Differentiate u to get du. Integrate dv to get v (no +C needed yet). Organise these in a small table: u on top-left, dv top-right, du bottom-left, v bottom-right.

Step 3
Apply the formula

Substitute into ∫u dv = uv − ∫v du. The new integral ∫v du should be simpler than the original.

Step 4
Evaluate and add +C

Evaluate ∫v du. Add +C only at the very end of the entire calculation, not after each step.

LIATE Rule (choosing u): When in doubt, choose u from the category that appears earliest in this list:
Logarithmic → Inverse trig → Algebraic (polynomials) → Trigonometric → Exponential
Choose u from the first category present. The remaining factor becomes dv.
§ 3 — Interactive Explorer

See the Formula in Action

Select any example below to see exactly how u, dv, du, and v are assigned, and how the formula is applied step by step.

🔢 Integration by Parts — Formula Explorer
The Integral
\(\int x e^x\,dx\)
u =
\(x\)
dv =
\(e^x\,dx\)
du =
\(dx\)
v =
\(e^x\)
Apply: ∫u·dv = uv − ∫v·du
u
\(x\)
·
v
\(e^x\)
−∫
v
\(e^x\)
·
du
\(dx\)
Result
\(\int x e^x\,dx = e^x(x-1)+C\)
💡 u = x (simpler derivative). dv = eˣdx (easy to integrate).
§ 4 — Worked Examples

Integration by Parts
— Fully Worked

Example 1 — $\int xe^{2x}\,dx$

A straightforward product of a polynomial and an exponential.

Choose: $u = x$ (algebraic), $dv = e^{2x}\,dx$ (exponential)

Compute: $du = dx$, $v = \dfrac{e^{2x}}{2}$

Apply formula:

$$\int xe^{2x}\,dx = x\cdot\frac{e^{2x}}{2} - \int\frac{e^{2x}}{2}\,dx = \frac{xe^{2x}}{2} - \frac{e^{2x}}{4} + C$$

$$= \boxed{\frac{e^{2x}(2x-1)}{4} + C}$$

Example 2 — $\int x^2\ln x\,dx$

A polynomial multiplied by a logarithm — logarithm always goes to u by LIATE.

Choose: $u = \ln x$ (logarithmic), $dv = x^2\,dx$

Compute: $du = \dfrac{1}{x}dx$, $v = \dfrac{x^3}{3}$

$$\int x^2\ln x\,dx = \frac{x^3}{3}\ln x - \int\frac{x^3}{3}\cdot\frac{1}{x}\,dx = \frac{x^3}{3}\ln x - \frac{1}{3}\int x^2\,dx$$

$$= \boxed{\frac{x^3}{3}\ln x - \frac{x^3}{9} + C}$$

Example 3 — $\int x^3 e^{x^2}\,dx$ (Substitution First!)

A case where you first do a substitution, then integration by parts.

Step 1 — Substitution: Let $t = x^2$, $dt = 2x\,dx$, so $x\,dx = \frac{dt}{2}$ and $x^3\,dx = x^2\cdot x\,dx = t\cdot\frac{dt}{2}$.

$$\int x^3 e^{x^2}\,dx = \frac{1}{2}\int t e^t\,dt$$

Step 2 — IBP on $\int te^t\,dt$: $u=t$, $dv=e^t\,dt$, $du=dt$, $v=e^t$.

$$\frac{1}{2}\int te^t\,dt = \frac{1}{2}(te^t - e^t) + C = \frac{1}{2}e^{x^2}(x^2-1)+C$$

$$= \boxed{\frac{e^{x^2}(x^2-1)}{2} + C}$$

Example 4 — $\int x^3 \ln^2 x\,dx$ (IBP Twice — Power meets Log²!)

A higher-degree polynomial paired with $\ln^2 x$ forces two full rounds of IBP with careful fraction tracking.

First IBP: $u = \ln^2 x$, $dv = x^3\,dx$, $du = \dfrac{2\ln x}{x}\,dx$, $v = \frac{x^4}{4}$.

$$\int x^3 \ln^2 x\,dx = \frac{x^4}{4}\ln^2 x - \frac{1}{2}\int x^3 \ln x\,dx \quad\cdots (*)$$

Second IBP on $\int x^3 \ln x\,dx$: $u = \ln x$, $dv = x^3\,dx$, $du = \frac{1}{x}\,dx$, $v = \frac{x^4}{4}$.

$$\int x^3 \ln x\,dx = \frac{x^4}{4}\ln x - \frac{1}{4}\int x^3\,dx = \frac{x^4}{4}\ln x - \frac{x^4}{16}$$

Substitute back into (*):

$$\int x^3 \ln^2 x\,dx = \frac{x^4}{4}\ln^2 x - \frac{1}{2}\!\left(\frac{x^4}{4}\ln x - \frac{x^4}{16}\right) + C$$

$$\boxed{\int x^3\ln^2 x\,dx = \frac{x^4}{4}\ln^2 x - \frac{x^4}{8}\ln x + \frac{x^4}{32} + C}$$

Example 5 — $\int (\ln x)^2\,dx$

Write $\int (\ln x)^2\cdot 1\,dx$ and use the LIATE trick.

IBP: $u = (\ln x)^2$, $dv = dx$, $du = \frac{2\ln x}{x}\,dx$, $v = x$.

$$\int(\ln x)^2\,dx = x(\ln x)^2 - 2\int\ln x\,dx$$

We know from Example 4 of the explorer that $\int\ln x\,dx = x\ln x - x + C$.

$$= x(\ln x)^2 - 2(x\ln x - x) + C = \boxed{x(\ln x)^2 - 2x\ln x + 2x + C}$$

📝 Important Note — Some Integrals Have Multiple Approaches

Consider $\int x^2 e^{x^3}\,dx$. You can solve it two ways:

Method 1 (Substitution): Let $u=x^3$, $du=3x^2\,dx$, so $x^2\,dx=\frac{du}{3}$. Then $\int x^2 e^{x^3}\,dx = \frac{1}{3}\int e^u\,du = \frac{e^{x^3}}{3}+C$. ✓ Easy!

Method 2 (IBP): $u=e^{x^3}$, $dv=x^2\,dx$, $du=3x^2e^{x^3}\,dx$, $v=\frac{x^3}{3}$. Then $\int x^2 e^{x^3}\,dx = \frac{x^3}{3}e^{x^3} - \int x^3 e^{x^3}\,dx$... which is harder than what we started with! ✗ IBP makes it worse.

The lesson: always try substitution first. If the integrand has a composite function with its derivative present (here $x^2$ is the derivative of $x^3$), substitution is the natural choice. IBP is for products where substitution won't simplify things.

Now consider $\int x\sqrt{x+5}\,dx$. Here both methods work cleanly — a great example to compare them side by side.

Method 1 (Substitution): Let $u = x+5$, so $x = u-5$ and $du = dx$.

$$\int x\sqrt{x+5}\,dx = \int (u-5)\sqrt{u}\,du = \int\!\left(u^{3/2} - 5u^{1/2}\right)du$$

$$= \frac{2}{5}u^{5/2} - \frac{10}{3}u^{3/2} + C = \frac{2}{5}(x+5)^{5/2} - \frac{10}{3}(x+5)^{3/2} + C$$

Method 2 (IBP): $u=x$, $dv=\sqrt{x+5}\,dx$, $du=dx$, $v=\frac{2}{3}(x+5)^{3/2}$.

$$\int x\sqrt{x+5}\,dx = \frac{2x}{3}(x+5)^{3/2} - \frac{2}{3}\int(x+5)^{3/2}\,dx$$

$$= \frac{2x}{3}(x+5)^{3/2} - \frac{2}{3}\cdot\frac{2}{5}(x+5)^{5/2} + C$$

$$= \frac{2x}{3}(x+5)^{3/2} - \frac{4}{15}(x+5)^{5/2} + C$$

Both answers are equivalent — you can verify by factoring out $(x+5)^{3/2}$. The lesson: when both methods work, substitution is usually shorter. IBP is powerful but requires more steps. Always choose the method that minimises algebra!

§ 5 — Definite Integration by Parts

Applying IBP to
Definite Integrals

Definite IBP Formula

For definite integrals, apply IBP and then evaluate at the limits:

$$\int_a^b u\,dv = \Big[uv\Big]_a^b - \int_a^b v\,du$$

Example 6 — $\int_0^1 xe^x\,dx$

IBP: $u=x$, $dv=e^x\,dx$, $du=dx$, $v=e^x$.

$$\int_0^1 xe^x\,dx = \Big[xe^x\Big]_0^1 - \int_0^1 e^x\,dx = (e-0) - \Big[e^x\Big]_0^1 = e - (e-1) = \boxed{1}$$

Example 7 — $\int_1^e x\ln x\,dx$

IBP: $u=\ln x$, $dv=x\,dx$, $du=\frac{dx}{x}$, $v=\frac{x^2}{2}$.

$$\int_1^e x\ln x\,dx = \left[\frac{x^2}{2}\ln x\right]_1^e - \int_1^e\frac{x}{2}\,dx = \frac{e^2}{2} - \left[\frac{x^2}{4}\right]_1^e = \frac{e^2}{2} - \frac{e^2-1}{4} = \boxed{\frac{e^2+1}{4}}$$

Example 8 — Area Under $y = xe^{-x}$ on $[0, 3]$

Find the area of the region bounded by $y = xe^{-x}$, the $x$-axis, $x=0$, and $x=3$.

Since $xe^{-x} \geq 0$ on $[0,3]$, area $= \int_0^3 xe^{-x}\,dx$.

IBP: $u=x$, $dv=e^{-x}\,dx$, $du=dx$, $v=-e^{-x}$.

$$= \Big[-xe^{-x}\Big]_0^3 + \int_0^3 e^{-x}\,dx = -3e^{-3} + \Big[-e^{-x}\Big]_0^3 = -3e^{-3} + (-e^{-3}+1) = 1-4e^{-3}$$

$$\approx \boxed{1 - 0.199 = 0.801 \text{ square units}}$$

Example 9 — Area Under $y = x\ln x$ on $[1, e]$

Find $\int_1^e x\ln x\,dx$ and interpret geometrically as the area under $y=x\ln x$ from $x=1$ to $x=e$.

Note: $x\ln x \geq 0$ for $x \geq 1$. From Example 7: area $= \dfrac{e^2+1}{4} \approx \dfrac{7.389+1}{4} \approx \mathbf{2.097}$ square units.

§ 6 — Integration by Parts More Than Once

When One Round of IBP
Is Not Enough

Sometimes after the first round of IBP, the new integral $\int v\,du$ still requires IBP again. This is common when the polynomial factor has degree 2 or higher.

Example 10 — $\int x^2 e^x\,dx$ (IBP Applied Twice)

First IBP: $u=x^2$, $dv=e^x\,dx$, $du=2x\,dx$, $v=e^x$.

$$\int x^2 e^x\,dx = x^2e^x - 2\int xe^x\,dx \quad\cdots(*)$$

Second IBP on $\int xe^x\,dx$: $u=x$, $dv=e^x\,dx$, $du=dx$, $v=e^x$.

$$\int xe^x\,dx = xe^x - e^x + C$$

Substitute into (*):

$$= x^2e^x - 2(xe^x - e^x) + C = \boxed{e^x(x^2-2x+2)+C}$$

Example 11 — $\int x^2\sin x\,dx$ (IBP Twice)

First IBP: $u=x^2$, $dv=\sin x\,dx$, $du=2x\,dx$, $v=-\cos x$.

$$\int x^2\sin x\,dx = -x^2\cos x + 2\int x\cos x\,dx$$

Second IBP on $\int x\cos x\,dx$: $u=x$, $dv=\cos x\,dx$, $du=dx$, $v=\sin x$.

$$\int x\cos x\,dx = x\sin x - \int\sin x\,dx = x\sin x + \cos x + C$$

$$\boxed{\int x^2\sin x\,dx = -x^2\cos x + 2x\sin x + 2\cos x + C}$$

Example 12 — $\int x^3\ln x\,dx$ (Single IBP)

Despite the higher power, only one IBP is needed because the logarithm differentiates to a simple fraction.

IBP: $u=\ln x$, $dv=x^3\,dx$, $du=\frac{dx}{x}$, $v=\frac{x^4}{4}$.

$$\int x^3\ln x\,dx = \frac{x^4}{4}\ln x - \int\frac{x^4}{4}\cdot\frac{1}{x}\,dx = \frac{x^4}{4}\ln x - \frac{1}{4}\int x^3\,dx = \boxed{\frac{x^4}{4}\ln x - \frac{x^4}{16}+C}$$

§ 7 — Business Applications

Integration by Parts
in Business & Economics

How to recognise when IBP is needed from a word problem: Look for expressions involving a polynomial multiplied by an exponential or logarithm — for example, a revenue rate like $f(t) = t \cdot e^{-rt}$ (linear growth times discounting factor). These always require IBP to evaluate the integral.

Example 13 — Present Value with Growing Income Rate

A Lahore-based tech startup generates income at the rate $f(t) = 200{,}000t$ PKR/year (revenue grows linearly). The annual interest rate is $r = 0.10$ compounded continuously. Find the present value of this income stream over $T = 3$ years.

$\text{PV} = \int_0^3 200{,}000t \cdot e^{-0.10t}\,dt = 200{,}000\int_0^3 te^{-0.10t}\,dt$

IBP: $u=t$, $dv=e^{-0.10t}\,dt$, $du=dt$, $v=-10e^{-0.10t}$.

$$\int_0^3 te^{-0.10t}\,dt = \Big[-10te^{-0.10t}\Big]_0^3 + 10\int_0^3 e^{-0.10t}\,dt = -30e^{-0.3} + 10\Big[-10e^{-0.10t}\Big]_0^3$$

$$= -30e^{-0.3} - 100(e^{-0.3}-1) = 100 - 130e^{-0.3} \approx 100 - 96.22 = 3.78$$

$$\text{PV} = 200{,}000 \times 3.78 \approx \boxed{\text{PKR }755{,}000}$$

Example 14 — Total Revenue from Marginal Revenue

A company's marginal revenue is $R'(x) = x\ln(x+1)$ PKR (hundred) per unit. Find the total revenue from selling the first 4 units.

$R(4)-R(0) = \int_0^4 x\ln(x+1)\,dx$

IBP: $u=\ln(x+1)$, $dv=x\,dx$, $du=\frac{dx}{x+1}$, $v=\frac{x^2}{2}$.

$$= \left[\frac{x^2}{2}\ln(x+1)\right]_0^4 - \int_0^4\frac{x^2}{2(x+1)}\,dx = 8\ln 5 - \frac{1}{2}\int_0^4\frac{x^2}{x+1}\,dx$$

For $\frac{x^2}{x+1}$, do polynomial division: $\frac{x^2}{x+1} = x-1+\frac{1}{x+1}$.

$$\frac{1}{2}\int_0^4\left(x-1+\frac{1}{x+1}\right)dx = \frac{1}{2}\left[\frac{x^2}{2}-x+\ln(x+1)\right]_0^4 = \frac{1}{2}(8-4+\ln 5) = 2+\frac{\ln 5}{2}$$

$$\text{Total revenue} = 8\ln 5 - 2 - \frac{\ln 5}{2} = \frac{15\ln 5}{2} - 2 \approx \boxed{\text{PKR }1{,}207}$$

Example 15 — Future Value with Linearly Growing Deposits

A Karachi factory deposits money continuously into an account at rate $f(t) = 50{,}000(1+t)$ PKR/year (deposits grow with time). The account earns $r = 8\%$ continuously. Find the future value after 2 years.

$\text{FV} = e^{0.16}\int_0^2 50{,}000(1+t)e^{-0.08t}\,dt$

Split: $\int_0^2 e^{-0.08t}\,dt + \int_0^2 te^{-0.08t}\,dt$

First integral: $\int_0^2 e^{-0.08t}\,dt = \left[\frac{e^{-0.08t}}{-0.08}\right]_0^2 = 12.5(1-e^{-0.16}) \approx 1.856$

Second (IBP): $u=t$, $dv=e^{-0.08t}\,dt$, $v=-12.5e^{-0.08t}$.

$$\int_0^2 te^{-0.08t}\,dt = \left[-12.5te^{-0.08t}\right]_0^2 + 12.5\int_0^2 e^{-0.08t}\,dt = -25e^{-0.16}+12.5(1.856) \approx 1.63$$

$$\text{FV} \approx 50{,}000 \times e^{0.16} \times (1.856+1.63) \approx 50{,}000\times 1.1735\times 3.486 \approx \boxed{\text{PKR }204{,}600}$$

Example 16 — Consumer's Surplus with Logarithmic Demand

The demand for a commodity is $D(q) = 10\ln(q+1) + 20$ PKR/unit, and the market price is set at $p_0 = D(4) = 10\ln 5 + 20 \approx 36.09$ PKR. Find the consumers' surplus.

$\text{CS} = \int_0^4[10\ln(q+1)+20]\,dq - p_0\cdot 4$

IBP for $\int\ln(q+1)\,dq$: $u=\ln(q+1)$, $dv=dq$, $du=\frac{dq}{q+1}$, $v=q$.

$$\int_0^4\ln(q+1)\,dq = \Big[q\ln(q+1)\Big]_0^4 - \int_0^4\frac{q}{q+1}\,dq = 4\ln 5 - \int_0^4\left(1-\frac{1}{q+1}\right)dq$$

$$= 4\ln 5 - [q-\ln(q+1)]_0^4 = 4\ln 5 - 4 + \ln 5 = 5\ln 5 - 4$$

$$\text{CS} = 10(5\ln 5-4)+80 - 4(10\ln 5+20) = 50\ln 5-40+80-40\ln 5-80 = 10\ln 5-40\approx\boxed{\text{PKR }-23.9}$$

Negative CS means consumers are paying more than their aggregate willingness to spend — this demand curve rises with quantity (unusual — normally demand curves slope down). In practice, always check that your demand function is decreasing for a sensible economic interpretation.

Example 17 — Net Profit with Advertising Spending

A company spends on advertising so that the rate of change of profit is $P'(t) = te^{0.5t}$ (PKR lakhs per month). Find the total increase in profit over the first 4 months.

$\Delta P = \int_0^4 te^{0.5t}\,dt$

IBP: $u=t$, $dv=e^{0.5t}\,dt$, $du=dt$, $v=2e^{0.5t}$.

$$= \Big[2te^{0.5t}\Big]_0^4 - 2\int_0^4 e^{0.5t}\,dt = 8e^2 - 2\Big[2e^{0.5t}\Big]_0^4 = 8e^2 - 4(e^2-1) = 4e^2+4$$

$$\approx 4(7.389)+4 = \boxed{\text{PKR }33.56 \text{ lakhs}}$$

§ 8 — Using Integral Tables

When IBP and Substitution
Are Not Enough

Most integrals in social and managerial sciences can be handled by substitution and integration by parts. But occasionally you encounter forms that require a lookup table. The key skill is recognising which form in the table your integral matches, and making the correct identification of the constants $a$, $b$, and $u$.

Note: Some integrals — like $\int \frac{e^x}{x}\,dx$ — cannot be evaluated by any method. No closed-form antiderivative exists. The integral table only lists those that can be evaluated.
Table 6.1 — A Short Table of Integrals
Forms Involving a + bu
1.$\displaystyle\int\frac{u}{a+bu}\,du = \frac{1}{b^2}\left[a+bu-a\ln|a+bu|\right]+C$
2.$\displaystyle\int\frac{u^2}{a+bu}\,du = \frac{1}{2b^3}\left[(a+bu)^2-4a(a+bu)+2a^2\ln|a+bu|\right]+C$
3.$\displaystyle\int\frac{u}{(a+bu)^2}\,du = \frac{1}{b^2}\left[\frac{a}{a+bu}+\ln|a+bu|\right]+C$
4.$\displaystyle\int\frac{u}{\sqrt{a+bu}}\,du = \frac{2}{3b^2}(bu-2a)\sqrt{a+bu}+C$
5.$\displaystyle\int\frac{du}{u\sqrt{a+bu}} = \frac{1}{\sqrt{a}}\ln\left|\frac{\sqrt{a+bu}-\sqrt{a}}{\sqrt{a+bu}+\sqrt{a}}\right|+C,\quad a>0$
6.$\displaystyle\int\frac{du}{u(a+bu)} = \frac{1}{a}\ln\left|\frac{u}{a+bu}\right|+C$
7.$\displaystyle\int\frac{du}{u^2(a+bu)} = -\frac{1}{au}+\frac{b}{a^2}\ln\left|\frac{a+bu}{u}\right|+C$
8.$\displaystyle\int\frac{du}{u^2(a+bu)^2} = -\frac{1}{a^2}\left[\frac{1}{u}+\frac{b}{a+bu}\right]+\frac{2b}{a^3}\ln\left|\frac{u}{a+bu}\right|+C$
Forms Involving a² + u²
9.$\displaystyle\int\sqrt{a^2+u^2}\,du = \frac{u}{2}\sqrt{a^2+u^2}+\frac{a^2}{2}\ln|u+\sqrt{a^2+u^2}|+C$
10.$\displaystyle\int\frac{du}{\sqrt{a^2+u^2}} = \ln|u+\sqrt{a^2+u^2}|+C$
11.$\displaystyle\int\frac{du}{u\sqrt{a^2+u^2}} = -\frac{1}{a}\ln\left|\frac{\sqrt{a^2+u^2}+a}{u}\right|+C$
12.$\displaystyle\int\frac{du}{(a^2+u^2)^{3/2}} = \frac{u}{a^2\sqrt{a^2+u^2}}+C$
13.$\displaystyle\int u^2\sqrt{a^2+u^2}\,du = \frac{u}{8}(2u^2+a^2)\sqrt{a^2+u^2}-\frac{a^4}{8}\ln|u+\sqrt{a^2+u^2}|+C$
Forms Involving a² − u²
14.$\displaystyle\int\frac{du}{u\sqrt{a^2-u^2}} = -\frac{1}{a}\ln\left|\frac{a+\sqrt{a^2-u^2}}{u}\right|+C$
15.$\displaystyle\int\frac{du}{u^2\sqrt{a^2-u^2}} = -\frac{\sqrt{a^2-u^2}}{a^2 u}+C$
16.$\displaystyle\int\frac{du}{a^2-u^2} = \frac{1}{2a}\ln\left|\frac{a+u}{a-u}\right|+C$
17.$\displaystyle\int\frac{\sqrt{a^2-u^2}}{u}\,du = \sqrt{a^2-u^2}-a\ln\left|\frac{a+\sqrt{a^2-u^2}}{u}\right|+C$
Forms Involving u² − a²
18.$\displaystyle\int\sqrt{u^2-a^2}\,du = \frac{u}{2}\sqrt{u^2-a^2}-\frac{a^2}{2}\ln|u+\sqrt{u^2-a^2}|+C$
19.$\displaystyle\int\frac{\sqrt{u^2-a^2}}{u^2}\,du = -\frac{\sqrt{u^2-a^2}}{u}+\ln|u+\sqrt{u^2-a^2}|+C$
20.$\displaystyle\int\frac{du}{\sqrt{u^2-a^2}} = \ln|u+\sqrt{u^2-a^2}|+C$
21.$\displaystyle\int\frac{du}{u^2\sqrt{u^2-a^2}} = \frac{\sqrt{u^2-a^2}}{a^2 u}+C$
Forms Involving eᵃᵘ and ln u
22.$\displaystyle\int ue^{au}\,du = \frac{1}{a^2}(au-1)e^{au}+C$
23.$\displaystyle\int\ln u\,du = u\ln u-u+C$
24.$\displaystyle\int\frac{du}{u\ln u} = \ln|\ln u|+C$
25.$\displaystyle\int u^n\ln u\,du = \frac{u^{n+1}}{n+1}\!\left(\ln u-\frac{1}{n+1}\right)+C,\quad n\neq -1$
Reduction Formulas
26.$\displaystyle\int u^n e^{au}\,du = \frac{1}{a}u^n e^{au}-\frac{n}{a}\int u^{n-1}e^{au}\,du$
27.$\displaystyle\int(\ln u)^n\,du = u(\ln u)^n-n\int(\ln u)^{n-1}\,du$
28.$\displaystyle\int u^n\sqrt{a+bu}\,du = \frac{2}{b(2n+3)}\!\left[u^n(a+bu)^{3/2}-na\int u^{n-1}\sqrt{a+bu}\,du\right],\quad n\neq-\tfrac{3}{2}$

Using the Table — Worked Examples

Example 18 — $\int\dfrac{1}{x(3+2x)}\,dx$

Match to Form 6: $\int\dfrac{du}{u(a+bu)} = \dfrac{1}{a}\ln\left|\dfrac{u}{a+bu}\right|+C$

Identify constants: $u=x$, $a=3$, $b=2$. The integrand matches Form 6 exactly — no substitution needed.

Apply Form 6 directly:

$$\int\frac{dx}{x(3+2x)} = \frac{1}{3}\ln\left|\frac{x}{3+2x}\right|+C$$

Verify by differentiating: $\frac{d}{dx}\left[\frac{1}{3}\ln|x| - \frac{1}{3}\ln|3+2x|\right] = \frac{1}{3x} - \frac{2}{3(3+2x)} = \frac{3+2x-2x}{3x(3+2x)} = \frac{1}{x(3+2x)}$ ✓

Example 19 — $\int\dfrac{1}{\sqrt{4x^2-9}}\,dx$

Match to Form 20: $\int\frac{du}{\sqrt{u^2-a^2}} = \ln|u+\sqrt{u^2-a^2}|+C$

Rewrite: $4x^2-9 = (2x)^2-3^2$. Let $u=2x$, $du=2\,dx$, $dx=\frac{du}{2}$, $a=3$.

$$\int\frac{dx}{\sqrt{4x^2-9}} = \frac{1}{2}\int\frac{du}{\sqrt{u^2-9}} = \frac{1}{2}\ln|u+\sqrt{u^2-9}|+C$$

$$= \boxed{\frac{1}{2}\ln|2x+\sqrt{4x^2-9}|+C}$$

Example 20 — $\int\dfrac{x}{(3+2x)^2}\,dx$

Match to Form 3 with $a=3$, $b=2$, $u=x$.

Form 3: $\int\frac{u}{(a+bu)^2}\,du = \frac{1}{b^2}\left[\frac{a}{a+bu}+\ln|a+bu|\right]+C$

With $a=3$, $b=2$:

$$\int\frac{x}{(3+2x)^2}\,dx = \frac{1}{4}\left[\frac{3}{3+2x}+\ln|3+2x|\right]+C$$

Example 21 — $\int x^3\ln x\,dx$ via Table

Match to Form 25 with $n=3$: $\int u^n\ln u\,du = \frac{u^{n+1}}{n+1}\left(\ln u - \frac{1}{n+1}\right)+C$

$$\int x^3\ln x\,dx = \frac{x^4}{4}\left(\ln x-\frac{1}{4}\right)+C = \boxed{\frac{x^4}{4}\ln x - \frac{x^4}{16}+C}$$

✓ Matches our IBP result from Example 12.

§ 9 — Logistic Equations

Logistic Growth:
When Growth Has a Ceiling

Simple exponential growth $\frac{dQ}{dt}=kQ$ says a population grows forever at the same rate — clearly unrealistic. The logistic model adds a carrying capacity $M$: growth slows as the population approaches $M$, and stops entirely once it reaches it.

Logistic equations model: the spread of COVID-19 in Pakistan, the adoption of new technology, the growth of a new business up to market saturation, and population dynamics of fish in a lake.

Logistic Differential Equation

A quantity $Q(t)$ satisfies the logistic equation if:

$$\frac{dQ}{dt} = kQ(M-Q)$$

where $k > 0$ is the growth rate constant and $M > 0$ is the carrying capacity (the maximum possible value of $Q$).

Solution: Separate variables and use partial fractions + Form 16 from the table $\int\frac{du}{a^2-u^2}$.

$$Q(t) = \frac{M}{1+Ae^{-kMt}}, \quad A = \frac{M-Q_0}{Q_0}$$

where $Q_0 = Q(0)$ is the initial value.

Example 22 — Solving a Logistic Equation

Solve $\frac{dQ}{dt} = 0.02Q(100-Q)$ with $Q(0)=10$.

Identify: $k=0.02$, $M=100$.

Apply the solution formula: $A = \frac{100-10}{10} = 9$, $kM = 0.02\times 100=2$.

$$\boxed{Q(t) = \frac{100}{1+9e^{-2t}}}$$

Check: $Q(0) = \frac{100}{1+9} = 10$ ✓. As $t\to\infty$, $Q(t)\to 100=M$ ✓.

Example 23 — Spread of Information in a Company

A company has 500 employees. A rumour spreads at rate $\frac{dQ}{dt} = 0.001Q(500-Q)$, where $Q$ is the number of people who have heard it. Initially 5 employees know. Find $Q(t)$ and determine when 250 employees will have heard it.

Solution: $k=0.001$, $M=500$, $Q_0=5$.

$A=\frac{500-5}{5}=99$, $kM=0.001\times 500=0.5$.

$$Q(t) = \frac{500}{1+99e^{-0.5t}}$$

When does $Q=250$?

$$250 = \frac{500}{1+99e^{-0.5t}} \Rightarrow 1+99e^{-0.5t}=2 \Rightarrow e^{-0.5t}=\frac{1}{99} \Rightarrow t = \frac{\ln 99}{0.5} \approx \boxed{9.2\text{ time units}}$$

Example 24 — Market Saturation of a New Product

A new mobile app in Pakistan has a potential market of $M = 2{,}000{,}000$ users. Market research suggests the spread follows $\frac{dQ}{dt} = 0.0000008\,Q(2{,}000{,}000 - Q)$, with $Q(0) = 500$ initial users. Find the number of users after 12 months.

$k=8\times 10^{-7}$, $M=2\times 10^6$, $Q_0=500$.

$A=\frac{2{,}000{,}000-500}{500}=3{,}999$, $kM=8\times10^{-7}\times 2\times10^6 = 1.6$.

$$Q(t)=\frac{2{,}000{,}000}{1+3{,}999\,e^{-1.6t}}$$

At $t=12$ months:

$$Q(12)=\frac{2{,}000{,}000}{1+3{,}999\,e^{-19.2}}\approx\frac{2{,}000{,}000}{1+3{,}999\times 4.5\times10^{-9}}\approx \boxed{\approx 1{,}999{,}982 \text{ users}}$$

After 12 months, essentially the entire potential market has adopted the app — rapid saturation due to the large $kM$ value.

Coming up next — §6.2 Numerical Integration — when even IBP and integral tables cannot give a closed-form answer, we compute definite integrals numerically using the Trapezoidal Rule and Simpson's Rule.
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