Integrals: 5 marks Questions (CBSE Class 12)

2 different 5 marks questions on Integrals from CBSE Class 12 Maths board exams 2026, newest first.

Find :
$\int \frac{x}{( x - 1 ) ( x ^{2} + 4 )} d x$

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Answer:

\frac{1}{5} lo g ∣ x - 1∣ - \frac{1}{10} lo g (x^{2} + 4) + \frac{2}{5} tan^{- 1} \frac{x}{2} + C

Let $\frac{x}{( x - 1 ) ( x ^{2} + 4 )} = \frac{A}{x - 1} + \frac{B x + C}{x ^{2} + 4}$ , so $x = A (x^{2} + 4) + (B x + C) (x - 1)$ .
$x = 1$ : $1 = 5 A \Rightarrow A = \frac{1}{5}$ . Coefficient of $x^{2}$ : $A + B = 0 \Rightarrow B = - \frac{1}{5}$ . Constant: $4 A - C = 0 \Rightarrow C = \frac{4}{5}$ .
Integral $= \frac{1}{5} \int \frac{d x}{x - 1} - \frac{1}{5} \int \frac{x d x}{x ^{2} + 4} + \frac{4}{5} \int \frac{d x}{x ^{2} + 4}$ .
$= \frac{1}{5} lo g ∣ x - 1∣ - \frac{1}{10} lo g (x^{2} + 4) + \frac{4}{5} \cdot \frac{1}{2} tan^{- 1} \frac{x}{2} + C$ .
$= \frac{1}{5} lo g ∣ x - 1∣ - \frac{1}{10} lo g (x^{2} + 4) + \frac{2}{5} tan^{- 1} \frac{x}{2} + C$ .

Q32 (OR) (OR)5 marksLong AnswerIntegrals CBSE 2026 · 65/4/1

Evaluate :
$\int_{0}^{1} \frac{x t a n ^{- 1} x}{( 1 + x ^{2} ) ^{3/2}} d x$

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Answer:

\frac{1}{2} (1 - \frac{π}{4}) = \frac{4 - π}{4 2}

Put $x = tan θ$ , $d x = sec^{2} θ d θ$ ; limits $0$ to $\frac{π}{4}$ ; $(1 + x^{2})^{3/2} = sec^{3} θ$ .
Integral $= \int_{0}^{π /4} \frac{θ t a n θ s e c ^{2} θ}{s e c ^{3} θ} d θ = \int_{0}^{π /4} θ sin θ d θ$ .
By parts: $= [- θ cos θ]_{0}^{π /4} + \int_{0}^{π /4} cos θ d θ = - \frac{π}{4} \cdot \frac{1}{2} + [sin θ]_{0}^{π /4}$ .
$= \frac{1}{2} - \frac{π}{4 2} = \frac{4 - π}{4 2}$ .

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