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Class 12Calculus6:26Published 16 Dec 2024

3 Board Questions on Integration

Three sure-shot Class XII board integration questions worked end to end, each using a different standard technique.

This lesson solves three integration problems of the kind that turn up on the Class XII board exam. The first uses long division and the rule for integrals that combine a function with its derivative times an exponential. The second is a clean substitution that lands on a logarithm, and the third uses integration by parts together with a tangent substitution.

What you'll learn

  • Recognising when an integrand fits the function-plus-its-derivative times an exponential pattern
  • Using long division to simplify a fraction before integrating it
  • Solving an integral by substitution to reach a logarithm
  • Applying integration by parts when one factor is a simple power of x

Lesson chapters

0:00Question 1: an exponential integral
0:54Dividing the fraction and splitting the terms
2:44Writing the second piece in derivative form
4:28Question 2: substitution to a logarithm
5:05Question 3: integration by parts with tangent

Lesson notes

This lesson works through three Class XII board questions on integration, each showing a standard technique you can reuse.

Question 1: an exponential integral

Evaluate

∫x2+1(x+1)2 ex dx.\int \frac{x^2 + 1}{(x+1)^2}\, e^{x}\, dx.

Whenever you see exe^{x} multiplying a fraction, look for the form

∫(f(x)+f′(x)) ex dx=f(x) ex+C.\int \big(f(x) + f'(x)\big)\, e^{x}\, dx = f(x)\, e^{x} + C.

Here the numerator and denominator are both degree 2, so first do a long division. Since (x+1)2=x2+2x+1(x+1)^2 = x^2 + 2x + 1,

x2+1x2+2x+1=1+−2x(x+1)2.\frac{x^2 + 1}{x^2 + 2x + 1} = 1 + \frac{-2x}{(x+1)^2}.

So the integral becomes

I=∫ex dx−2∫x(x+1)2 ex dx=ex−2∫x(x+1)2 ex dx.I = \int e^{x}\, dx - 2\int \frac{x}{(x+1)^2}\, e^{x}\, dx = e^{x} - 2\int \frac{x}{(x+1)^2}\, e^{x}\, dx.

Writing the second piece in derivative form

Add and subtract 11 in the numerator: x=(x+1)−1x = (x+1) - 1, so

x(x+1)2=1x+1−1(x+1)2.\frac{x}{(x+1)^2} = \frac{1}{x+1} - \frac{1}{(x+1)^2}.

Now take f(x)=1x+1f(x) = \dfrac{1}{x+1}. Then f′(x)=−1(x+1)2f'(x) = -\dfrac{1}{(x+1)^2}, which is exactly the f(x)+f′(x)f(x) + f'(x) pattern, so

∫(1x+1−1(x+1)2)ex dx=exx+1.\int \left( \frac{1}{x+1} - \frac{1}{(x+1)^2} \right) e^{x}\, dx = \frac{e^{x}}{x+1}.

Putting it together,

I=ex−2exx+1+C.I = e^{x} - \frac{2e^{x}}{x+1} + C.

Question 2: substitution to a logarithm

Evaluate

I=∫3axb2+c2x2 dx.I = \int \frac{3ax}{b^2 + c^2 x^2}\, dx.

Let t=b2+c2x2t = b^2 + c^2 x^2, so dt=2c2x dxdt = 2c^2 x\, dx, which gives x dx=12c2 dtx\, dx = \dfrac{1}{2c^2}\, dt. Then

I=3a2c2∫1t dt=3a2c2ln⁡∣t∣+C=3a2c2ln⁡∣b2+c2x2∣+C.I = \frac{3a}{2c^2} \int \frac{1}{t}\, dt = \frac{3a}{2c^2} \ln|t| + C = \frac{3a}{2c^2} \ln\left| b^2 + c^2 x^2 \right| + C.

Question 3: integration by parts with tangent

Evaluate

I=∫x tan⁡x sec⁡2x dx.I = \int x\, \tan x\, \sec^2 x\, dx.

Take the first function as xx and the second as tan⁡x sec⁡2x\tan x\, \sec^2 x. For the second function, substitute t=tan⁡xt = \tan x, so dt=sec⁡2x dxdt = \sec^2 x\, dx and

∫tan⁡x sec⁡2x dx=∫t dt=t22=tan⁡2x2.\int \tan x\, \sec^2 x\, dx = \int t\, dt = \frac{t^2}{2} = \frac{\tan^2 x}{2}.

Applying integration by parts,

I=x⋅tan⁡2x2−12∫tan⁡2x dx.I = x \cdot \frac{\tan^2 x}{2} - \frac{1}{2}\int \tan^2 x\, dx.

Using tan⁡2x=sec⁡2x−1\tan^2 x = \sec^2 x - 1,

∫tan⁡2x dx=tan⁡x−x.\int \tan^2 x\, dx = \tan x - x.

Therefore

I=x2tan⁡2x−12tan⁡x+x2+C.I = \frac{x}{2}\tan^2 x - \frac{1}{2}\tan x + \frac{x}{2} + C.

Key takeaways

  • If exe^{x} multiplies a fraction, try to reach the form ∫(f+f′)ex dx=f ex+C\int (f + f')e^{x}\, dx = f\, e^{x} + C, dividing first when the degrees match.
  • A factor of xx next to a quadratic denominator often signals the substitution t=t = (the quadratic), leading to a logarithm.
  • Integration by parts with xx as the first function reduces the problem to a simpler integral you can finish with a trig identity.