Find : (1)/(5+4x-x²)dx

MathsCBSE Class 12 · Board exam 2024, set 65/5/1 · 2 marks · Very short answer

CBSE Class 12 Mathematics 2024: question 24(b)

Set in the CBSE Class 12 board exam of 2024, paper 65/5/1, section B, question 24(b), for 2 marks. Try it first; the hints and our step-by-step solution are below.

Q24(b) · 2 marks · Very short answerIntegrals

Find : $\int \frac{1}{5 + 4 x - x ^{2}} d x$

Source: official question paper from cbse.gov.in (PDF). See it in the full 2024 board paper, set 65/5/1.

Hints

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

Hint 1

Complete the square in $- x^{2} + 4 x + 5$ .

Hint 2

The result has the form $a^{2} - u^{2}$ ; recall the standard integral of $\frac{1}{a ^{2} - u ^{2}}$ .

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.

Complete the square: $5 + 4 x - x^{2} = 9 - (x^{2} - 4 x + 4) = 3^{2} - (x - 2)^{2}$ , so $I = \int \frac{d x}{3 ^{2} - ( x - 2 ) ^{2}}$ .
1 mark
Using $\int \frac{d u}{a ^{2} - u ^{2}} = \frac{1}{2 a} lo g \frac{a + u}{a - u} + C$ : $I = \frac{1}{6} lo g \frac{3 + x - 2}{3 - x + 2} + C = \frac{1}{6} lo g \frac{1 + x}{5 - x} + C$ .
1 mark

Answer: $\frac{1}{6} lo g \frac{1 + x}{5 - x} + C$

Where the marks are

Writes denominator as $3^{2} - (x - 2)^{2}$ (1 mark)
Applies standard formula to get $\frac{1}{6} lo g \frac{1 + x}{5 - x} + C$ (1 mark)

Revise the chapter

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

Practise this chapterBoard-style questions from Integrals, one at a time, marked against the answer key and marking points.Back to the full paper2024 board paper, set 65/5/1: every question, by section.

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CBSE Class 12 Mathematics 2024: question 24(b)

Hints

Step-by-step solution

Revise the chapter

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