Official Academic Study Notes • Published: August 24, 2026 • Free Printable Resource
1. Topic Introduction & Key Concepts
Integration (also known as anti-differentiation) is the fundamental process of finding a function given its derivative. In FSc Part 2 Mathematics (Chapter 3), integration serves as a core pillar of Calculus, heavily weighted in both FBISE and Punjab Board examinations.
Fundamental Rules & Theorems of Integration
- Power Rule: $\int x^n dx = \frac{x^{n+1}}{n+1} + C \quad (n \neq -1)$
- Extended Power Rule: $\int [f(x)]^n f'(x) dx = \frac{[f(x)]^{n+1}}{n+1} + C \quad (n \neq -1)$
- Logarithmic Rule: $\int \frac{f'(x)}{f(x)} dx = \ln|f(x)| + C$
- Exponential Integrals: $\int e^{ax} dx = \frac{e^{ax}}{a} + C$ and $\int a^{kx} dx = \frac{a^{kx}}{k \ln a} + C$
- Trigonometric Integrals:$$\int \sin(ax) dx = -\frac{\cos(ax)}{a} + C, \quad \int \cos(ax) dx = \frac{\sin(ax)}{a} + C$$$\int \sec^2 x dx = \tan x + C, \quad \int \csc^2 x dx = -\cot x + C$$$\int \sec x \tan x dx = \sec x + C, \quad \int \csc x \cot x dx = -\csc x + C$$
- Special Integrals (Inverse Trigonometric):$$\int \frac{1}{\sqrt{a^2 - x^2}} dx = \sin^{-1}\left(\frac{x}{a}\right) + C, \quad \int \frac{1}{a^2 + x^2} dx = \frac{1}{a} \tan^{-1}\left(\frac{x}{a}\right) + C$$
- Integration by Parts Formula:$$\int u \cdot v \, dx = u \int v \, dx - \int \left( \frac{du}{dx} \cdot \int v \, dx \right) dx$$
- Special Exponential Theorem:$$\int e^{ax} [a f(x) + f'(x)] dx = e^{ax} f(x) + C$$
2. Step-by-Step Solved Important Questions
Question 1: Evaluate $\int \frac{1}{x^2 + 4x + 13} dx$
Solution:
Step 1: Complete the square in the denominator.
$$x^2 + 4x + 13 = (x^2 + 4x + 4) + 9 = (x + 2)^2 + 3^2$$
Step 2: Substitute this into the integral:
$$I = \int \frac{1}{(x + 2)^2 + 3^2} dx$$
Step 3: Apply the standard formula $\int \frac{1}{x^2 + a^2} dx = \frac{1}{a} \tan^{-1}\left(\frac{x}{a}\right) + C$ where $u = x + 2$ and $a = 3$:
$$I = \frac{1}{3} \tan^{-1}\left(\frac{x + 2}{3}\right) + C$$
Question 2: Evaluate $\int x \tan^{-1}x \, dx$ using Integration by Parts
Solution:
Step 1: Choose $u$ and $v$ using the ILATE rule. Let $u = \tan^{-1}x$ (Inverse Trig) and $v = x$ (Algebraic).
Step 2: Apply the Integration by Parts formula:
$$\int u \cdot v \, dx = u \int v \, dx - \int \left( \frac{d}{dx}[u] \int v \, dx \right) dx$$
$$I = \tan^{-1}x \left(\frac{x^2}{2}\right) - \int \left( \frac{1}{1 + x^2} \cdot \frac{x^2}{2} \right) dx$$
$$I = \frac{x^2}{2} \tan^{-1}x - \frac{1}{2} \int \frac{x^2}{1 + x^2} dx$$
Step 3: Rewrite $\frac{x^2}{1 + x^2}$ as $\frac{(1 + x^2) - 1}{1 + x^2} = 1 - \frac{1}{1 + x^2}$:
$$I = \frac{x^2}{2} \tan^{-1}x - \frac{1}{2} \int \left(1 - \frac{1}{1 + x^2}\right) dx$$
$$I = \frac{x^2}{2} \tan^{-1}x - \frac{1}{2} \left[ x - \tan^{-1}x \right] + C$$
$$I = \frac{x^2}{2} \tan^{-1}x - \frac{x}{2} + \frac{1}{2} \tan^{-1}x + C = \frac{1}{2}(x^2 + 1) \tan^{-1}x - \frac{x}{2} + C$$
Question 3: Evaluate $\int e^x \left( \frac{1 + \sin x}{1 + \cos x} \right) dx$
Solution:
Step 1: Express trigonometric terms using half-angle identities:
$$1 + \sin x = 1 + 2\sin\left(\frac{x}{2}\right)\cos\left(\frac{x}{2}\right)$$
$$1 + \cos x = 2\cos^2\left(\frac{x}{2}\right)$$
Step 2: Rewrite the integrand:
$$\frac{1 + \sin x}{1 + \cos x} = \frac{1 + 2\sin\left(\frac{x}{2}\right)\cos\left(\frac{x}{2}\right)}{2\cos^2\left(\frac{x}{2}\right)} = \frac{1}{2\cos^2\left(\frac{x}{2}\right)} + \frac{2\sin\left(\frac{x}{2}\right)\cos\left(\frac{x}{2}\right)}{2\cos^2\left(\frac{x}{2}\right)}$$
$$= \frac{1}{2} \sec^2\left(\frac{x}{2}\right) + \tan\left(\frac{x}{2}\right)$$
Step 3: Compare with theorem $\int e^{ax}[a f(x) + f'(x)] dx = e^{ax} f(x) + C$ where $a=1$:
Let $f(x) = \tan\left(\frac{x}{2}\right) \implies f'(x) = \frac{1}{2} \sec^2\left(\frac{x}{2}\right)$.
$$I = \int e^x \left[ \tan\left(\frac{x}{2}\right) + \frac{1}{2}\sec^2\left(\frac{x}{2}\right) \right] dx = e^x \tan\left(\frac{x}{2}\right) + C$$
Question 4: Evaluate the Definite Integral $\int_{0}^{\frac{\pi}{4}} \frac{1}{1 + \sin x} dx$
Solution:
Step 1: Multiply numerator and denominator by $(1 - \sin x)$:
$$I = \int_{0}^{\frac{\pi}{4}} \frac{1 - \sin x}{(1 + \sin x)(1 - \sin x)} dx = \int_{0}^{\frac{\pi}{4}} \frac{1 - \sin x}{1 - \sin^2 x} dx = \int_{0}^{\frac{\pi}{4}} \frac{1 - \sin x}{\cos^2 x} dx$$
Step 2: Split into two terms:
$$I = \int_{0}^{\frac{\pi}{4}} \left( \frac{1}{\cos^2 x} - \frac{\sin x}{\cos^2 x} \right) dx = \int_{0}^{\frac{\pi}{4}} \left( \sec^2 x - \sec x \tan x \right) dx$$
Step 3: Integrate and substitute upper and lower limits:
$$I = \left[ \tan x - \sec x \right]_{0}^{\frac{\pi}{4}}$$
$$I = \left( \tan\frac{\pi}{4} - \sec\frac{\pi}{4} \right) - (\tan 0 - \sec 0)$$
$$I = (1 - \sqrt{2}) - (0 - 1) = 2 - \sqrt{2}$$
3. Solved Important Board MCQs
- 1. Evaluate $\int \frac{f'(x)}{\sqrt{f(x)}} dx$:
A) $\ln|f(x)| + C$
B) $2\sqrt{f(x)} + C$
C) $\sqrt{f(x)} + C$
D) $\frac{1}{2\sqrt{f(x)}} + C$
Rationale: Rewrite as $\int [f(x)]^{-1/2} f'(x) dx$. By the extended power rule, $\frac{[f(x)]^{1/2}}{1/2} + C = 2\sqrt{f(x)} + C$. - 2. What is $\int e^{3x} (3 \sin x + \cos x) dx$?
A) $e^{3x} \cos x + C$
B) $e^{3x} \sin x + C$
C) $3e^{3x} \sin x + C$
D) $-e^{3x} \cos x + C$
Rationale: Using $\int e^{ax}[a f(x) + f'(x)]dx = e^{ax}f(x) + C$ with $a=3$ and $f(x) = \sin x$, $f'(x) = \cos x$. - 3. $\int \sec x \, dx = ?$
A) $\tan x + C$
B) $\ln|\sec x + \tan x| + C$
C) $\ln|\sec x - \tan x| + C$
D) $\sec x \tan x + C$
Rationale: Standard direct integration formula for $\sec x$. - 4. Evaluate $\int_{-2}^{2} x^3 \cos x \, dx$:
A) $8$
B) $0$
C) $4$
D) $2$
Rationale: The integrand $g(x) = x^3 \cos x$ is an odd function because $g(-x) = (-x)^3 \cos(-x) = -x^3 \cos x = -g(x)$. The integral of an odd function over symmetric limits $[-a, a]$ is always $0$. - 5. $\int \frac{1}{x \ln x} dx$ is equal to:
A) $\ln x + C$
B) $\ln|\ln x| + C$
C) $(\ln x)^2 + C$
D) $\frac{1}{(\ln x)^2} + C$
Rationale: Rewrite as $\int \frac{1/x}{\ln x} dx$. Since the numerator is the derivative of denominator $f(x) = \ln x$, the integral equals $\ln|\ln x| + C$. - 6. $\int a^{x} dx = ?$
A) $a^x + C$
B) $a^x \ln a + C$
C) $\frac{a^x}{\ln a} + C$
D) $\frac{a^{x+1}}{x+1} + C$
Rationale: Direct standard derivative relation: $\frac{d}{dx}(a^x) = a^x \ln a$, hence $\int a^x dx = \frac{a^x}{\ln a} + C$. - 7. The value of $\int_{1}^{e} \frac{1}{x} dx$ is:
A) $0$
B) $1$
C) $e$
D) $\frac{1}{e}$
Rationale: $\int_{1}^{e} \frac{1}{x} dx = [\ln x]_{1}^{e} = \ln(e) - \ln(1) = 1 - 0 = 1$. - 8. $\int \frac{1}{\sqrt{1-x^2}} dx = ?$
A) $\sin^{-1}x + C$
B) $\cos^{-1}x + C$
C) $\tan^{-1}x + C$
D) $\sec^{-1}x + C$
Rationale: Standard derivative of $\sin^{-1}x$ is $\frac{1}{\sqrt{1-x^2}}$, so its integral is $\sin^{-1}x + C$.