Master indefinite and definite integrals — substitution, partial fractions, integration by parts, Fundamental Theorem of Calculus — with visual diagrams, step-marked solutions, MCQs, A-R, and PYQs.
Understanding integration
Differential Calculus is centred on the concept of the derivative. Integral Calculus is motivated by the problem of defining and calculating the area of the region bounded by the graph of functions.
If a function $f$ is differentiable in an interval I, then the functions that could possibly have $f$ as a derivative are called antiderivatives (or primitive) of the function.
The formula that gives all these antiderivatives is called the indefinite integral of the function, and the process of finding antiderivatives is called integration.
Integration is the inverse process of differentiation.
Here, $C$ is called the constant of integration (an arbitrary real number).
$\{F + C, C \in \mathbb{R}\}$ denotes a family of antiderivatives of $f$.
Remark: Functions with the same derivative differ by a constant.
| Symbol / Term | Meaning |
|---|---|
| $\int f(x)dx$ | Integral of $f$ with respect to $x$ |
| $f(x)$ in $\int f(x)dx$ | Integrand |
| $x$ in $\int f(x)dx$ | Variable of integration |
| Integrate | Find the integral |
| An integral of $f$ | A function $F$ such that $F'(x) = f(x)$ |
| Integration | The process of finding the integral |
| Constant of Integration | Any real number $C$, considered as constant function |
See integration concepts in action
Essential formulae to memorise
Basic rules of integration
$$\frac{d}{dx}\int f(x)dx = f(x) \quad \text{and} \quad \int f'(x)dx = f(x) + C$$
Two indefinite integrals with the same derivative lead to the same family of curves and so they are equivalent.
$$\int [f(x) + g(x)]dx = \int f(x)dx + \int g(x)dx$$
For any real number $k$: $\int k f(x)dx = k \int f(x)dx$
$$\int [k_1f_1(x) + \ldots + k_nf_n(x)]dx = k_1\int f_1 + \ldots + k_n\int f_n$$
Substitution, partial fractions, by parts
If $x = g(t)$, then $dx = g'(t)dt$ and:
Common identities used:
| Form of Rational Function | Partial Fraction Form |
|---|---|
| $\frac{px+q}{(x-a)(x-b)}$ | $\frac{A}{x-a} + \frac{B}{x-b}$ |
| $\frac{px+q}{(x-a)^2}$ | $\frac{A}{x-a} + \frac{B}{(x-a)^2}$ |
| $\frac{px^2+qx+r}{(x-a)(x-b)(x-c)}$ | $\frac{A}{x-a}+\frac{B}{x-b}+\frac{C}{x-c}$ |
| $\frac{px^2+qx+r}{(x-a)^2(x-b)}$ | $\frac{A}{x-a}+\frac{B}{(x-a)^2}+\frac{C}{x-b}$ |
| $\frac{px^2+qx+r}{(x-a)(x^2+bx+c)}$ | $\frac{A}{x-a}+\frac{Bx+C}{x^2+bx+c}$ |
Choose first function using ILATE:
$$\int e^x[f(x) + f'(x)]dx = e^x f(x) + C$$
Standard special forms
Area under the curve and evaluation
$\int_a^b f(x)dx$ is called the definite integral of $f$ over the range $[a,b]$, where $a$ is the lower limit and $b$ is the upper limit.
It represents the area of the region bounded by the curve $y = f(x)$, the ordinates $x=a$, $x=b$, and the x-axis.
Key Idea: When you substitute, also change the limits!
The bridge between differentiation and integration
If $f$ is continuous on $[a,b]$, then the area function $A(x)$ is defined as:
$$A(x) = \int_a^x f(t)dt$$
Let $f$ be continuous on $[a,b]$ and $A(x) = \int_a^x f(t)dt$. Then:
Interpretation: The derivative of the area function is the function itself.
Let $f$ be continuous on $[a,b]$ and $F$ be an antiderivative of $f$. Then:
Why "Fundamental"? It gives an easy way to compute definite integrals using antiderivatives, without computing limits of sums!
Powerful tools for evaluation
$$\int_a^b f(x)dx = \int_a^b f(t)dt$$
$$\int_a^b f(x)dx = -\int_b^a f(x)dx ; \quad \int_a^a f(x)dx = 0$$
$$\int_a^b f(x)dx = \int_a^c f(x)dx + \int_c^b f(x)dx$$
$$\int_a^b f(x)dx = \int_a^b f(a+b-x)dx$$
$$\int_0^a f(x)dx = \int_0^a f(a-x)dx$$
$$\int_0^{2a} f(x)dx = \int_0^a f(x)dx + \int_0^a f(2a-x)dx$$
$\int_0^{2a} f(x)dx = 2\int_0^a f(x)dx$ if $f(2a-x) = f(x)$
$\int_0^{2a} f(x)dx = 0$ if $f(2a-x) = -f(x)$
(i) If $f$ is even: $\int_{-a}^{a} f(x)dx = 2\int_0^a f(x)dx$
(ii) If $f$ is odd: $\int_{-a}^{a} f(x)dx = 0$
$$\int_0^a f(x)dx = \int_0^a f(a-x)dx$$
Replace $x$ by $(a-x)$ in the integrand. Great for trigonometric integrals.
Before computing, check if $f(-x) = f(x)$ (even) or $f(-x) = -f(x)$ (odd).
For odd: integral = 0. For even: double the integral from 0 to a.
Speed techniques for integration
"Is the derivative of some part present?"
Order: I → L → A → T → E
ax² + bx + c = a[(x + b/2a)² + (c/a − b²/4a²)]
Before computing $\int_{-a}^a f(x)dx$, check symmetry!
$\int_0^a f(x)dx = \int_0^a f(a-x)dx$
Degree of numerator ≥ denominator? First divide!
Change the limits too!
Simplify before integrating
$\int e^x[f(x) + f'(x)]dx = e^x f(x) + C$
Always verify:
Click "Show Answer" to reveal step-by-step solutions with marks
Q1. Find $\int \frac{x^3 - 1}{x^2}dx$
$\frac{x^3-1}{x^2} = x - x^{-2}$
$\int(x - x^{-2})dx = \frac{x^2}{2} - \frac{x^{-1}}{-1} + C = \frac{x^2}{2} + \frac{1}{x} + C$
$\frac{x^2}{2} + \frac{1}{x} + C$
Q2. Find $\int \sin^3 x \cos^2 x \, dx$
$\sin^3 x\cos^2 x = \sin^2 x \cos^2 x \sin x = (1-\cos^2 x)\cos^2 x \sin x$
Put $t = \cos x$, so $dt = -\sin x dx$
$\int (1-t^2)t^2(-dt) = \int (t^4 - t^2)dt$
$= \frac{t^5}{5} - \frac{t^3}{3} + C = \frac{\cos^5 x}{5} - \frac{\cos^3 x}{3} + C$
Q3. Find $\int \frac{dx}{(x+1)(x+2)}$
$\frac{1}{(x+1)(x+2)} = \frac{A}{x+1} + \frac{B}{x+2}$
$1 = A(x+2) + B(x+1)$
Put $x=-1$: $A = 1$ ; Put $x=-2$: $B = -1$
$\int\left(\frac{1}{x+1} - \frac{1}{x+2}\right)dx = \log|x+1| - \log|x+2| + C$
$= \log\left|\frac{x+1}{x+2}\right| + C$
Q4. Find $\int x e^x dx$
By ILATE: $x$ is Algebraic → first function; $e^x$ is Exponential → second function
$\int x e^x dx = x \int e^x dx - \int (1)(\int e^x dx)dx$
$= x e^x - \int e^x dx$
$= x e^x - e^x + C = e^x(x-1) + C$
$e^x(x-1) + C$
Q5. Evaluate $\int_2^3 x^2 dx$
$\int x^2 dx = \frac{x^3}{3} = F(x)$
$\int_2^3 x^2 dx = F(3) - F(2) = \frac{27}{3} - \frac{8}{3}$
$= \frac{19}{3}$
Q6. Evaluate $\int_0^{\pi/2} \frac{\sin x}{\sin x + \cos x}dx$
Let $I = \int_0^{\pi/2} \frac{\sin x}{\sin x + \cos x}dx$
By P₄: $I = \int_0^{\pi/2} \frac{\sin(\pi/2-x)}{\sin(\pi/2-x) + \cos(\pi/2-x)}dx = \int_0^{\pi/2} \frac{\cos x}{\cos x + \sin x}dx$
$2I = \int_0^{\pi/2} \frac{\sin x + \cos x}{\sin x + \cos x}dx = \int_0^{\pi/2} 1\,dx = \frac{\pi}{2}$
$I = \frac{\pi}{4}$
Q7. Evaluate $\int_{-1}^{1} x^3 \cos^2 x \, dx$
Let $f(x) = x^3\cos^2 x$
$f(-x) = (-x)^3\cos^2(-x) = -x^3\cos^2 x = -f(x)$
$f$ is odd
For odd function: $\int_{-a}^a f(x)dx = 0$
Therefore, $\int_{-1}^{1} x^3\cos^2 x\,dx = 0$
Answer is 0
Select your answer and click "Check Answers"
Each question has an Assertion (A) and a Reason (R). Choose the correct option.
(A) Both A and R are true, and R is the correct explanation of A.
(B) Both A and R are true, but R is NOT the correct explanation of A.
(C) A is true, but R is false.
(D) A is false, but R is true.
Frequently asked questions in board exams
Q. Find $\int \frac{x^3 - 1}{x^2}dx$
$\frac{x^3-1}{x^2} = x - x^{-2}$
$\int(x - x^{-2})dx = \frac{x^2}{2} + \frac{1}{x} + C$
$\frac{x^2}{2} + \frac{1}{x} + C$
Q. Find $\int \sin^3 x \cos^2 x \, dx$
$\sin^3 x\cos^2 x = (1-\cos^2 x)\cos^2 x \sin x$
$t = \cos x$, $dt = -\sin x dx$
$\int (1-t^2)t^2(-dt) = \int (t^4 - t^2)dt$
$= \frac{\cos^5 x}{5} - \frac{\cos^3 x}{3} + C$
Q. Evaluate $\int_0^{\pi/2} \frac{\sin x}{\sin x + \cos x}dx$
$I = \int_0^{\pi/2}\frac{\cos x}{\cos x + \sin x}dx$
$2I = \int_0^{\pi/2}1\,dx = \frac{\pi}{2}$
$I = \frac{\pi}{4}$
Q. Find $\int x e^x dx$
First function = $x$ (Algebraic), Second = $e^x$ (Exponential)
$= x e^x - \int 1 \cdot e^x dx$
$= e^x(x-1) + C$
$e^x(x-1) + C$
Q. Evaluate $\int_0^1 x e^{x^2}dx$
Put $t = x^2$, $dt = 2x dx$
x=0 → t=0, x=1 → t=1
$= \frac{1}{2}\int_0^1 e^t dt = \frac{1}{2}[e^t]_0^1$
$= \frac{e-1}{2}$
Q. Find $\int \frac{dx}{(x+1)(x+2)}$
$\frac{1}{(x+1)(x+2)} = \frac{A}{x+1} + \frac{B}{x+2}$
$1 = A(x+2) + B(x+1)$
$A = 1, B = -1$
$= \int\left(\frac{1}{x+1}-\frac{1}{x+2}\right)dx = \log\left|\frac{x+1}{x+2}\right| + C$
$\log\left|\frac{x+1}{x+2}\right| + C$
Q. Evaluate $\int_{-1}^{1} x^3 \cos^2 x \, dx$
$f(x) = x^3\cos^2 x$
$f(-x) = -x^3\cos^2 x = -f(x)$
$f$ is odd
For odd function: integral = 0
$\int_{-1}^{1} x^3\cos^2 x \, dx = 0$
Q. Evaluate $\int_2^3 x^2 dx$
$\int x^2 dx = \frac{x^3}{3}$
$= \frac{27}{3} - \frac{8}{3}$
$= \frac{19}{3}$