Find : √((x)/(1-x³))dx; x(0,1).

MathsCBSE Class 12 · Sample paper, session 2023-24 · 3 marks · Short answer

CBSE Class 12 Mathematics sample paper 2023-24: question 28(a)

From the official CBSE sample question paper for 2023-24, section C, question 28(a), for 3 marks. Try it first; the hints and our step-by-step solution are below.

Q28(a) · 3 marks · Short answerIntegrals

Find : $\int \frac{x}{1 - x ^{3}} d x$ ; $x \in (0, 1)$ .

Source: official sample paper from cbseacademic.nic.in (PDF). See it in the full Sample paper 2023-24.

Hints

Open one hint at a time and try again before the next.

Hint 1

Write $x^{3}$ as $(x^{3/2})^{2}$ .

Hint 2

Find the derivative of $x^{3/2}$ and compare it with the $x$ in the numerator.

Step-by-step solution

Each step shows the marks it earns, as an examiner would award them. Checked by an independent re-solve from the question paper.

Let $x^{3/2} = t \Rightarrow d t = \frac{3}{2} x^{1/2} d x \Rightarrow x d x = \frac{2}{3} d t$ .
½ mark
$\int \frac{x}{1 - x ^{3}} d x = \frac{2}{3} \int \frac{d t}{1 - t ^{2}}$ .
½ mark
$= \frac{2}{3} sin^{- 1} t + C$ .
1 mark
$= \frac{2}{3} sin^{- 1} (x^{3/2}) + C$ .
1 mark

Answer: $\frac{2}{3} sin^{- 1} (x^{3/2}) + C$

Where the marks are

Chooses the substitution $t = x^{3/2}$ (½ mark)
Converts to $\frac{2}{3} \int \frac{d t}{1 - t ^{2}}$ (½ mark)
Integrates to $\frac{2}{3} sin^{- 1} t$ (1 mark)
Back-substitutes to give the answer in $x$ (1 mark)

Revise the chapter

Chapter: Integrals (notes, formulas, where marks are lost and every past question)
Topic: Integration by Substitution

Practise this chapterBoard-style questions from Integrals, one at a time, marked against the answer key and marking points.Back to the full paperSample paper 2023-24: every question, by section.

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CBSE Class 12 Mathematics sample paper 2023-24: question 28(a)

Hints

Step-by-step solution

Revise the chapter

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