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NCERT Solutions · Class 12 Mathematics Integrals

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EXERCISE 7.4 1–10 (part 10 of 27)

  1. Integrate the functions in Exercises $\displaystyle 1$ to 23.

    Exercise 1

    3x2x6+1\displaystyle \frac{3 x^{2}}{x^{6}+1}

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    NCERT’s answer
    \(\displaystyle \tan ^{-1} x^{3}+\mathrm{C}\)
    The integrand becomes a standard form under a substitution, using \(\displaystyle \int \frac{dt}{t^{2}+1}=\tan^{-1}t+\mathrm{C}\).Put \(\displaystyle t=x^{3}\). Then \(\displaystyle dt=3x^{2}\,dx\), which is exactly the numerator, and \(\displaystyle x^{6}=\left(x^{3}\right)^{2}=t^{2}\). \[\int \frac{3x^{2}}{x^{6}+1}\,dx=\int \frac{dt}{t^{2}+1}=\tan^{-1}t+\mathrm{C}\] Restoring \(\displaystyle t=x^{3}\), \[\int \frac{3x^{2}}{x^{6}+1}\,dx=\tan^{-1}\left(x^{3}\right)+\mathrm{C}\]

  2. Exercise 2

    11+4x2\displaystyle \frac{1}{\sqrt{1+4 x^{2}}}

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    NCERT’s answer
    \(\displaystyle \frac{1}{2} \log \left|2 x+\sqrt{1+4 x^{2}}\right|+\mathrm{C}\)
    Standard form: \(\displaystyle \int \frac{dt}{\sqrt{t^{2}+a^{2}}}=\log\left|t+\sqrt{t^{2}+a^{2}}\right|+\mathrm{C}\).Write \(\displaystyle 1+4x^{2}=1+(2x)^{2}\) and put \(\displaystyle t=2x\), so \(\displaystyle dt=2\,dx\), i.e. \(\displaystyle dx=\frac{dt}{2}\). \[\int \frac{dx}{\sqrt{1+4x^{2}}}=\frac{1}{2}\int \frac{dt}{\sqrt{t^{2}+1^{2}}}=\frac{1}{2}\log\left|t+\sqrt{t^{2}+1}\right|+\mathrm{C}\] Hence \[\int \frac{dx}{\sqrt{1+4x^{2}}}=\frac{1}{2}\log\left|2x+\sqrt{1+4x^{2}}\right|+\mathrm{C}\]

  3. Exercise 3

    1(2−x)2+1\displaystyle \frac{1}{\sqrt{(2-x)^{2}+1}}

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    NCERT’s answer
    \(\displaystyle \log \left|\frac{1}{2-x+\sqrt{x^{2}-4 x+5}}\right|+\mathrm{C}\)
    Standard form: \(\displaystyle \int \frac{dt}{\sqrt{t^{2}+a^{2}}}=\log\left|t+\sqrt{t^{2}+a^{2}}\right|+\mathrm{C}\).Put \(\displaystyle t=2-x\). Then \(\displaystyle dt=-\,dx\), so \(\displaystyle dx=-\,dt\) — this minus sign is the step most often dropped here. \[\int \frac{dx}{\sqrt{(2-x)^{2}+1}}=-\int \frac{dt}{\sqrt{t^{2}+1^{2}}}=-\log\left|t+\sqrt{t^{2}+1}\right|+\mathrm{C}\] Since \(\displaystyle (2-x)^{2}+1=x^{2}-4x+5\), \[\int \frac{dx}{\sqrt{(2-x)^{2}+1}}=-\log\left|(2-x)+\sqrt{x^{2}-4x+5}\right|+\mathrm{C}\]

  4. Exercise 4

    19−25x2\displaystyle \frac{1}{\sqrt{9-25 x^{2}}}

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    NCERT’s answer
    \(\displaystyle \frac{1}{5} \sin ^{-1} \frac{5 x}{3}+C\)
    Standard form: \(\displaystyle \int \frac{dx}{\sqrt{a^{2}-x^{2}}}=\sin^{-1}\frac{x}{a}+\mathrm{C}\).The coefficient of \(\displaystyle x^{2}\) must first be made \(\displaystyle 1\) inside the radical: \[9-25x^{2}=25\left(\frac{9}{25}-x^{2}\right)\quad\Rightarrow\quad \sqrt{9-25x^{2}}=5\sqrt{\left(\frac{3}{5}\right)^{2}-x^{2}}\] Therefore \[\int \frac{dx}{\sqrt{9-25x^{2}}}=\frac{1}{5}\int \frac{dx}{\sqrt{\left(\dfrac{3}{5}\right)^{2}-x^{2}}}=\frac{1}{5}\sin^{-1}\!\left(\frac{x}{3/5}\right)+\mathrm{C}\] \[=\frac{1}{5}\sin^{-1}\!\left(\frac{5x}{3}\right)+\mathrm{C},\qquad |x|<\frac{3}{5}\]

  5. Exercise 5

    3x1+2x4\displaystyle \frac{3 x}{1+2 x^{4}}

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    NCERT’s answer
    \(\displaystyle \frac{3}{2 \sqrt{2}} \tan ^{-1} \sqrt{2} x^{2}+\mathrm{C}\)
    Substitute to turn the quartic into a quadratic, then use \(\displaystyle \int \frac{dt}{t^{2}+a^{2}}=\frac{1}{a}\tan^{-1}\frac{t}{a}+\mathrm{C}\).Put \(\displaystyle t=x^{2}\). Then \(\displaystyle dt=2x\,dx\), so \(\displaystyle x\,dx=\frac{dt}{2}\), and \(\displaystyle x^{4}=t^{2}\). \[\int \frac{3x}{1+2x^{4}}\,dx=\frac{3}{2}\int \frac{dt}{1+2t^{2}}=\frac{3}{4}\int \frac{dt}{t^{2}+\left(\dfrac{1}{\sqrt{2}}\right)^{2}}\] (the factor \(\displaystyle 2\) in the denominator is taken outside, which is where the \(\displaystyle \frac{3}{4}\) comes from). \[=\frac{3}{4}\cdot\frac{1}{1/\sqrt{2}}\tan^{-1}\!\left(\frac{t}{1/\sqrt{2}}\right)+\mathrm{C}=\frac{3\sqrt{2}}{4}\tan^{-1}\left(\sqrt{2}\,t\right)+\mathrm{C}\] Hence \[\int \frac{3x}{1+2x^{4}}\,dx=\frac{3}{2\sqrt{2}}\tan^{-1}\left(\sqrt{2}\,x^{2}\right)+\mathrm{C}\]

  6. Exercise 6

    x21−x6\displaystyle \frac{x^{2}}{1-x^{6}}

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    NCERT’s answer
    \(\displaystyle \frac{1}{6} \log \left|\frac{1+x^{3}}{1-x^{3}}\right|+\mathrm{C}\)
    Standard form: \(\displaystyle \int \frac{dt}{1-t^{2}}=\frac{1}{2}\log\left|\frac{1+t}{1-t}\right|+\mathrm{C}\).Put \(\displaystyle t=x^{3}\). Then \(\displaystyle dt=3x^{2}\,dx\), so \(\displaystyle x^{2}\,dx=\frac{dt}{3}\), and \(\displaystyle x^{6}=t^{2}\). \[\int \frac{x^{2}}{1-x^{6}}\,dx=\frac{1}{3}\int \frac{dt}{1-t^{2}}=\frac{1}{3}\cdot\frac{1}{2}\log\left|\frac{1+t}{1-t}\right|+\mathrm{C}\] Hence \[\int \frac{x^{2}}{1-x^{6}}\,dx=\frac{1}{6}\log\left|\frac{1+x^{3}}{1-x^{3}}\right|+\mathrm{C}\]

  7. Exercise 7

    x−1x2−1\displaystyle \frac{x-1}{\sqrt{x^{2}-1}}

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    NCERT’s answer
    \(\displaystyle \sqrt{x^{2}-1}-\log \left|x+\sqrt{x^{2}-1}\right|+C\)
    Split the numerator so that one piece is a multiple of the derivative of \(\displaystyle x^{2}-1\) and the other is a standard form. \[\int \frac{x-1}{\sqrt{x^{2}-1}}\,dx=\int \frac{x}{\sqrt{x^{2}-1}}\,dx-\int \frac{dx}{\sqrt{x^{2}-1}}\] For the first integral put \(\displaystyle u=x^{2}-1\), so \(\displaystyle du=2x\,dx\): \[\int \frac{x\,dx}{\sqrt{x^{2}-1}}=\frac{1}{2}\int u^{-1/2}\,du=\sqrt{u}=\sqrt{x^{2}-1}\] The second is the standard form \(\displaystyle \int \frac{dx}{\sqrt{x^{2}-a^{2}}}=\log\left|x+\sqrt{x^{2}-a^{2}}\right|\) with \(\displaystyle a=1\).Therefore \[\int \frac{x-1}{\sqrt{x^{2}-1}}\,dx=\sqrt{x^{2}-1}-\log\left|x+\sqrt{x^{2}-1}\right|+\mathrm{C}\]

  8. Exercise 8

    x2x6+a6\displaystyle \frac{x^{2}}{\sqrt{x^{6}+a^{6}}}

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    NCERT’s answer
    \(\displaystyle \frac{1}{3} \log \left|x^{3}+\sqrt{x^{6}+a^{6}}\right|+\mathrm{C}\)
    Standard form: \(\displaystyle \int \frac{dt}{\sqrt{t^{2}+a^{2}}}=\log\left|t+\sqrt{t^{2}+a^{2}}\right|+\mathrm{C}\).Put \(\displaystyle t=x^{3}\), so \(\displaystyle dt=3x^{2}\,dx\), i.e. \(\displaystyle x^{2}\,dx=\frac{dt}{3}\). Also \(\displaystyle x^{6}=t^{2}\) and \(\displaystyle a^{6}=\left(a^{3}\right)^{2}\). \[\int \frac{x^{2}}{\sqrt{x^{6}+a^{6}}}\,dx=\frac{1}{3}\int \frac{dt}{\sqrt{t^{2}+\left(a^{3}\right)^{2}}}=\frac{1}{3}\log\left|t+\sqrt{t^{2}+a^{6}}\right|+\mathrm{C}\] Hence \[\int \frac{x^{2}}{\sqrt{x^{6}+a^{6}}}\,dx=\frac{1}{3}\log\left|x^{3}+\sqrt{x^{6}+a^{6}}\right|+\mathrm{C}\]

  9. Exercise 9

    sec⁡2xtan⁡2x+4\displaystyle \frac{\sec ^{2} x}{\sqrt{\tan ^{2} x+4}}

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    NCERT’s answer
    \(\displaystyle \log \left|\tan x+\sqrt{\tan ^{2} x+4}\right|+C\)
    The numerator \(\displaystyle \sec^{2}x\) is the derivative of \(\displaystyle \tan x\), so substitute for \(\displaystyle \tan x\).Put \(\displaystyle t=\tan x\). Then \(\displaystyle dt=\sec^{2}x\,dx\). \[\int \frac{\sec^{2}x}{\sqrt{\tan^{2}x+4}}\,dx=\int \frac{dt}{\sqrt{t^{2}+2^{2}}}\] By \(\displaystyle \int \frac{dt}{\sqrt{t^{2}+a^{2}}}=\log\left|t+\sqrt{t^{2}+a^{2}}\right|+\mathrm{C}\) with \(\displaystyle a=2\), \[=\log\left|t+\sqrt{t^{2}+4}\right|+\mathrm{C}=\log\left|\tan x+\sqrt{\tan^{2}x+4}\right|+\mathrm{C}\]

  10. Exercise 10

    1x2+2x+2\displaystyle \frac{1}{\sqrt{x^{2}+2 x+2}}

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    NCERT’s answer
    \(\displaystyle \log \left|x+1+\sqrt{x^{2}+2 x+2}\right|+\mathrm{C}\)
    Complete the square in the radicand, then apply \(\displaystyle \int \frac{dt}{\sqrt{t^{2}+a^{2}}}=\log\left|t+\sqrt{t^{2}+a^{2}}\right|+\mathrm{C}\). \[x^{2}+2x+2=\left(x^{2}+2x+1\right)+1=(x+1)^{2}+1^{2}\] Put \(\displaystyle t=x+1\), so \(\displaystyle dt=dx\). \[\int \frac{dx}{\sqrt{x^{2}+2x+2}}=\int \frac{dt}{\sqrt{t^{2}+1}}=\log\left|t+\sqrt{t^{2}+1}\right|+\mathrm{C}\] Hence \[\int \frac{dx}{\sqrt{x^{2}+2x+2}}=\log\left|(x+1)+\sqrt{x^{2}+2x+2}\right|+\mathrm{C}\]