Exercise 7.4, Question 1
\(\displaystyle\int \dfrac{3x^2}{x^6 + 1}\,dx\)
Show solution
- \(t = x^3\), \(dt = 3x^2\,dx\): \(\int\dfrac{dt}{t^2 + 1}\).
Answer: \(\tan^{-1}(x^3) + C\)
NCERT Solutions for Class 12 Maths Chapter 7 Exercise 7.4
Exercise 7.4: Integrals of some particular functions. Standard forms: \(\int\dfrac{dx}{x^2 - a^2} = \dfrac{1}{2a}\log\left|\dfrac{x - a}{x + a}\right|\), \(\int\dfrac{dx}{a^2 - x^2} = \dfrac{1}{2a}\log\left|\dfrac{a + x}{a - x}\right|\), \(\int\dfrac{dx}{x^2 + a^2} = \dfrac1a\tan^{-1}\dfrac{x}{a}\), \(\int\dfrac{dx}{\sqrt{x^2 \pm a^2}} = \log\left|x + \sqrt{x^2 \pm a^2}\right|\), \(\int\dfrac{dx}{\sqrt{a^2 - x^2}} = \sin^{-1}\dfrac{x}{a}\). Complete the square in a quadratic; for \(\dfrac{px + q}{\text{quadratic}}\) write \(px + q = A\,(\text{quadratic})' + B\).
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Exercise 7.4 questions and solutions
Exercise 7.4, Question 2
\(\displaystyle\int \dfrac{1}{\sqrt{1 + 4x^2}}\,dx\)
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- \(t = 2x\), \(dx = \tfrac12dt\): \(\tfrac12\int\dfrac{dt}{\sqrt{1 + t^2}}\).
Answer: \[\tfrac12\log\left|2x + \sqrt{1 + 4x^2}\right| + C\]
Exercise 7.4, Question 3
\(\displaystyle\int \dfrac{1}{\sqrt{(2 - x)^2 + 1}}\,dx\)
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- \(t = 2 - x\), \(dx = -dt\): \[-\int\dfrac{dt}{\sqrt{t^2 + 1}} = -\log\left|t + \sqrt{t^2 + 1}\right|\]
Answer: \[-\log\left|2 - x + \sqrt{x^2 - 4x + 5}\right| + C\]
Exercise 7.4, Question 4
\(\displaystyle\int \dfrac{1}{\sqrt{9 - 25x^2}}\,dx\)
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- \(t = 5x\): \(\tfrac15\int\dfrac{dt}{\sqrt{3^2 - t^2}}\).
Answer: \(\tfrac15\sin^{-1}\dfrac{5x}{3} + C\)
Exercise 7.4, Question 5
\(\displaystyle\int \dfrac{3x}{1 + 2x^4}\,dx\)
Show solution
- \(t = \sqrt2\,x^2\), \(dt = 2\sqrt2\,x\,dx\): \[\dfrac{3}{2\sqrt2}\int\dfrac{dt}{1 + t^2}\]
Answer: \[\dfrac{3}{2\sqrt2}\tan^{-1}(\sqrt2\,x^2) + C\]
Exercise 7.4, Question 6
\(\displaystyle\int \dfrac{x^2}{1 - x^6}\,dx\)
Show solution
- \(t = x^3\): \[\tfrac13\int\dfrac{dt}{1 - t^2} = \tfrac13 \cdot \tfrac12\log\left|\dfrac{1 + t}{1 - t}\right|\]
Answer: \[\tfrac16\log\left|\dfrac{1 + x^3}{1 - x^3}\right| + C\]
Exercise 7.4, Question 7
\(\displaystyle\int \dfrac{x - 1}{\sqrt{x^2 - 1}}\,dx\)
Show solution
- Split: \[\dfrac{x}{\sqrt{x^2 - 1}} - \dfrac{1}{\sqrt{x^2 - 1}}\]; the first gives \(\sqrt{x^2 - 1}\) (put \(t = x^2 - 1\)).
Answer: \[\sqrt{x^2 - 1} - \log\left|x + \sqrt{x^2 - 1}\right| + C\]
Exercise 7.4, Question 8
\(\displaystyle\int \dfrac{x^2}{\sqrt{x^6 + a^6}}\,dx\)
Show solution
- \(t = x^3\): \[\tfrac13\int\dfrac{dt}{\sqrt{t^2 + (a^3)^2}}\]
Answer: \[\tfrac13\log\left|x^3 + \sqrt{x^6 + a^6}\right| + C\]
Exercise 7.4, Question 9
\(\displaystyle\int \dfrac{\sec^2 x}{\sqrt{\tan^2 x + 4}}\,dx\)
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- \(t = \tan x\), \(dt = \sec^2 x\,dx\): \(\int\dfrac{dt}{\sqrt{t^2 + 2^2}}\).
Answer: \[\log\left|\tan x + \sqrt{\tan^2 x + 4}\right| + C\]
Exercise 7.4, Question 10
\(\displaystyle\int \dfrac{1}{\sqrt{x^2 + 2x + 2}}\,dx\)
Show solution
- \(x^2 + 2x + 2 = (x + 1)^2 + 1\).
Answer: \[\log\left|x + 1 + \sqrt{x^2 + 2x + 2}\right| + C\]
Exercise 7.4, Question 11
\(\displaystyle\int \dfrac{1}{9x^2 + 6x + 5}\,dx\)
Show solution
- \(9x^2 + 6x + 5 = (3x + 1)^2 + 4\); \(t = 3x + 1\): \(\tfrac13\int\dfrac{dt}{t^2 + 2^2}\).
Answer: \(\tfrac16\tan^{-1}\dfrac{3x + 1}{2} + C\)
Exercise 7.4, Question 12
\(\displaystyle\int \dfrac{1}{\sqrt{7 - 6x - x^2}}\,dx\)
Show solution
- \(7 - 6x - x^2 = 16 - (x + 3)^2\).
Answer: \(\sin^{-1}\dfrac{x + 3}{4} + C\)
Exercise 7.4, Question 13
\(\displaystyle\int \dfrac{1}{\sqrt{(x - 1)(x - 2)}}\,dx\)
Show solution
- \[\begin{aligned}(x - 1)(x - 2) &= x^2 - 3x + 2 \\ &= \left(x - \tfrac32\right)^2 - \tfrac14\end{aligned}\]
Answer: \[\log\left|x - \tfrac32 + \sqrt{x^2 - 3x + 2}\right| + C\]
Exercise 7.4, Question 14
\(\displaystyle\int \dfrac{1}{\sqrt{8 + 3x - x^2}}\,dx\)
Show solution
- \[8 + 3x - x^2 = \tfrac{41}{4} - \left(x - \tfrac32\right)^2\]
Answer: \(\sin^{-1}\dfrac{2x - 3}{\sqrt{41}} + C\)
Exercise 7.4, Question 15
\(\displaystyle\int \dfrac{1}{\sqrt{(x - a)(x - b)}}\,dx\)
Show solution
- \[(x - a)(x - b) = \left(x - \dfrac{a + b}{2}\right)^2 - \left(\dfrac{a - b}{2}\right)^2\]
Answer: \[\log\left|x - \dfrac{a + b}{2} + \sqrt{(x - a)(x - b)}\right| + C\]
Exercise 7.4, Question 16
\(\displaystyle\int \dfrac{4x + 1}{\sqrt{2x^2 + x - 3}}\,dx\)
Show solution
- \(4x + 1\) is the derivative of \(2x^2 + x - 3\): \(\int t^{-1/2}\,dt\).
Answer: \(2\sqrt{2x^2 + x - 3} + C\)
Exercise 7.4, Question 17
\(\displaystyle\int \dfrac{x + 2}{\sqrt{x^2 - 1}}\,dx\)
Show solution
- \[\dfrac{x}{\sqrt{x^2 - 1}} + \dfrac{2}{\sqrt{x^2 - 1}}\]
Answer: \[\sqrt{x^2 - 1} + 2\log\left|x + \sqrt{x^2 - 1}\right| + C\]
Exercise 7.4, Question 18
\(\displaystyle\int \dfrac{5x - 2}{1 + 2x + 3x^2}\,dx\)
Show solution
- \[5x - 2 = \tfrac56(6x + 2) - \tfrac{11}{3}\]
- \[\begin{aligned}\int\dfrac{dx}{3x^2 + 2x + 1} &= \tfrac13\int\dfrac{dx}{\left(x + \frac13\right)^2 + \frac29} \\ &= \tfrac13 \cdot \tfrac{3}{\sqrt2}\tan^{-1}\dfrac{3x + 1}{\sqrt2}\end{aligned}\]
Answer: \[\tfrac56\log|3x^2 + 2x + 1| - \dfrac{11}{3\sqrt2}\tan^{-1}\dfrac{3x + 1}{\sqrt2} + C\]
Exercise 7.4, Question 19
\(\displaystyle\int \dfrac{6x + 7}{\sqrt{(x - 5)(x - 4)}}\,dx\)
Show solution
- \(6x + 7 = 3(2x - 9) + 34\), and \[x^2 - 9x + 20 = \left(x - \tfrac92\right)^2 - \tfrac14\]
Answer: \[6\sqrt{x^2 - 9x + 20} + 34\log\left|x - \tfrac92 + \sqrt{x^2 - 9x + 20}\right| + C\]
Exercise 7.4, Question 20
\(\displaystyle\int \dfrac{x + 2}{\sqrt{4x - x^2}}\,dx\)
Show solution
- \(x + 2 = -\tfrac12(4 - 2x) + 4\); \(4x - x^2 = 4 - (x - 2)^2\).
Answer: \[-\sqrt{4x - x^2} + 4\sin^{-1}\dfrac{x - 2}{2} + C\]
Exercise 7.4, Question 21
\(\displaystyle\int \dfrac{x + 2}{\sqrt{x^2 + 2x + 3}}\,dx\)
Show solution
- \(x + 2 = \tfrac12(2x + 2) + 1\); \(x^2 + 2x + 3 = (x + 1)^2 + 2\).
Answer: \[\sqrt{x^2 + 2x + 3} + \log\left|x + 1 + \sqrt{x^2 + 2x + 3}\right| + C\]
Exercise 7.4, Question 22
\(\displaystyle\int \dfrac{x + 3}{x^2 - 2x - 5}\,dx\)
Show solution
- \(x + 3 = \tfrac12(2x - 2) + 4\); \(x^2 - 2x - 5 = (x - 1)^2 - (\sqrt6)^2\).
- \[4\int\dfrac{dx}{(x - 1)^2 - 6} = \dfrac{4}{2\sqrt6}\log\left|\dfrac{x - 1 - \sqrt6}{x - 1 + \sqrt6}\right|\]
Answer: \[\tfrac12\log|x^2 - 2x - 5| + \dfrac{2}{\sqrt6}\log\left|\dfrac{x - 1 - \sqrt6}{x - 1 + \sqrt6}\right| + C\]
Exercise 7.4, Question 23
\(\displaystyle\int \dfrac{5x + 3}{\sqrt{x^2 + 4x + 10}}\,dx\)
Show solution
- \(5x + 3 = \tfrac52(2x + 4) - 7\); \(x^2 + 4x + 10 = (x + 2)^2 + 6\).
Answer: \[5\sqrt{x^2 + 4x + 10} - 7\log\left|x + 2 + \sqrt{x^2 + 4x + 10}\right| + C\]
Exercise 7.4, Question 24
\(\displaystyle\int\dfrac{dx}{x^2 + 2x + 2}\) equals: (A) \(x\tan^{-1}(x + 1) + C\) (B) \(\tan^{-1}(x + 1) + C\) (C) \((x + 1)\tan^{-1}x + C\) (D) \(\tan^{-1}x + C\)
Show solution
- \(x^2 + 2x + 2 = (x + 1)^2 + 1\).
Answer: (B)
Exercise 7.4, Question 25
\(\displaystyle\int\dfrac{dx}{\sqrt{9x - 4x^2}}\) equals: (A) \(\tfrac19\sin^{-1}\dfrac{9x - 8}{8} + C\) (B) \(\tfrac12\sin^{-1}\dfrac{8x - 9}{9} + C\) (C) \(\tfrac13\sin^{-1}\dfrac{9x - 8}{8} + C\) (D) \(\tfrac12\sin^{-1}\dfrac{9x - 8}{9} + C\)
Show solution
- \[9x - 4x^2 = 4\left[\tfrac{81}{64} - \left(x - \tfrac98\right)^2\right]\]
- \[\tfrac12\int\dfrac{dx}{\sqrt{\left(\frac98\right)^2 - \left(x - \frac98\right)^2}} = \tfrac12\sin^{-1}\dfrac{x - \frac98}{\frac98}\]
Answer: (B)
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