Indefinite Integration — Class 12 IMO

Anti-derivatives and Standard Forms

Integration reverses differentiation. An antiderivative (indefinite integral) of $f$ is a function $F$ with $F'(x)=f(x)$; we write $\displaystyle\int f(x)\,dx=F(x)+C$, where $C$ is the arbitrary constant of integration.

Standard integrals

$\displaystyle\int x^{n}\,dx=\frac{x^{n+1}}{n+1}+C\ (n\ne-1)$, $\ \displaystyle\int \frac1x\,dx=\ln|x|+C$, $\ \displaystyle\int e^{x}\,dx=e^{x}+C$, $\ \displaystyle\int \cos x\,dx=\sin x+C$, $\ \displaystyle\int \sec^2x\,dx=\tan x+C$.

Three core techniques

Substitution: if the integrand contains a function and its derivative, put $u=g(x)$, $du=g'(x)\,dx$.
Integration by parts: $\displaystyle\int u\,dv=uv-\int v\,du$. Choose $u$ by the ILATE order (Inverse, Log, Algebraic, Trig, Exponential).
Partial fractions: split a proper rational function into simpler fractions before integrating.

Example 1: Evaluate $\displaystyle\int (3x^2+2x)\,dx$.

$=3\cdot\dfrac{x^3}{3}+2\cdot\dfrac{x^2}{2}+C=x^3+x^2+C.$

Example 2: Evaluate $\displaystyle\int 2x\,e^{x^2}\,dx$.

Put $u=x^2$, $du=2x\,dx$. Then $\int e^{u}\,du=e^{u}+C=e^{x^2}+C.$

Example 3: Evaluate $\displaystyle\int x\,e^{x}\,dx$.

By parts with $u=x,\ dv=e^x dx$: $u=x,\ du=dx,\ v=e^x$. $\int xe^x dx=xe^x-\int e^x dx=xe^x-e^x+C=e^x(x-1)+C.$

Example 4: Evaluate $\displaystyle\int \frac{1}{x^2-1}\,dx$.

Partial fractions: $\dfrac{1}{x^2-1}=\dfrac{1}{2}\left(\dfrac{1}{x-1}-\dfrac{1}{x+1}\right)$. Integrating, $=\dfrac12\ln\left|\dfrac{x-1}{x+1}\right|+C.$

Quick recap

$\int f\,dx=F+C$ where $F'=f$; never omit the constant $C$.
$\int x^n dx=\dfrac{x^{n+1}}{n+1}+C$ ($n\ne-1$); $\int\tfrac1x dx=\ln|x|+C$.
Substitution targets "function and its derivative" patterns.
By parts: $\int u\,dv=uv-\int v\,du$, choose $u$ via ILATE.
Partial fractions break proper rational integrands into simple terms.

✓ Quick check

Find ∫ x sin x dx

By parts: u = x, dv = sin x dx. du = dx, v = -cos x. ∫ u dv = uv - ∫ v du = -x cos x - ∫ (-cos x) dx = -x cos x + sin x + C.

Evaluate: ∫ 1/√(1 - x²) dx

This is a standard fundamental formula. The derivative of sin⁻¹ x is 1/√(1 - x²).

Substitution and Partial Fractions

Substitution sets u = g(x) so that du absorbs g′(x)dx, turning the integral into a standard form.

Partial fractions split a rational function into simpler fractions before integrating.

Example 1: Find ∫2x·e^(x²) dx.

Let u = x²: e^(x²) + C.

Example 2: ∫1/((x−1)(x−2)) dx splits using which method?

Partial fractions.

Quick recap

Substitution: choose u whose derivative appears.
Partial fractions simplify rational integrands.

✓ Quick check

∫dx/√(x² + 9) is equal to:

∫dx/√(x² + a²) = ln|x + √(x² + a²)| + C or sinh⁻¹(x/a) + C. Both forms are equivalent.

Evaluate: ∫ dx / (3 + 2x - x²)

Complete square: 3 + 2x - x² = 4 - (x-1)². Integral is ∫ dx / (2² - (x-1)²). Formula gives (1/(2*2)) log|(2+(x-1)) / (2-(x-1))| = (1/4) log|(x+1)/(3-x)| + C.

Integration by Parts

Integration by parts: ∫u dv = uv − ∫v du. Choose u by the ILATE order (Inverse, Log, Algebraic, Trig, Exponential).

It is ideal for products such as x·eˣ, x·sin x and ln x.

Example 1: Find ∫x·eˣ dx.

x·eˣ − eˣ + C = eˣ(x − 1) + C.

Example 2: Find ∫ln x dx.

x ln x − x + C.

Quick recap

∫u dv = uv − ∫v du.
Pick u by ILATE order.

✓ Quick check

The rate of depreciation of a machine in a factory is dV/dt = -500 (in ₹/year). If the current value (t=0) is ₹ 1,00,000, what is its value after 10 years?

V(t) = ∫ -500 dt = -500t + C. V(0) = 100000. At t=10, V(10) = -5000 + 100000 = 95000.

In a market, the marginal cost function for producing x kg of sweets is MC = 5 + 2x. The cost of producing 10 kg is ₹200. The total cost function C(x) is:

C(x) = ∫MC dx = ∫(5+2x)dx = 5x + x² + K. Given C(10) = 50 + 100 + K = 200 → K = 50. So C(x) = 5x + x² + 50.