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NCERT Solutions · Class 12 Mathematics Integrals

261 questions · 261 still being checked

EXERCISE 7.2 11–20 (part 4 of 27)

  1. Integrate the functions in Exercises $\displaystyle 1$ to $\displaystyle 37$:

    Exercise 11

    xx+4,x>0\displaystyle \frac{x}{\sqrt{x+4}}, x>0

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    NCERT’s answer
    \(\displaystyle \frac{2}{3} \sqrt{x+4}(x-8)+\mathrm{C}\)
    Put \[t=x+4\quad\Rightarrow\quad x=t-4,\quad dx=dt,\] so that (with \(\displaystyle x>0\), hence \(\displaystyle t>4>0\)) \[\int\frac{x}{\sqrt{x+4}}\,dx=\int\frac{t-4}{\sqrt{t}}\,dt=\int\left(t^{1/2}-4t^{-1/2}\right)dt=\frac{2}{3}t^{3/2}-8t^{1/2}+\mathrm{C}.\] Resubstituting \(\displaystyle t=x+4\), \[\int\frac{x}{\sqrt{x+4}}\,dx=\frac{2}{3}(x+4)^{3/2}-8\sqrt{x+4}+\mathrm{C}=\frac{2}{3}\sqrt{x+4}\,(x-8)+\mathrm{C}.\]

  2. Exercise 12

    (x3−1)13x5\displaystyle \left(x^{3}-1\right)^{\frac{1}{3}} x^{5}

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    NCERT’s answer
    \(\displaystyle \frac{1}{7}\left(x^{3}-1\right)^{\frac{7}{3}}+\frac{1}{4}\left(x^{3}-1\right)^{\frac{4}{3}}+\mathrm{C}\)
    Split the power of \(\displaystyle x\) so that one part supplies the differential: \(\displaystyle x^{5}\,dx=x^{3}\cdot x^{2}\,dx\). Put \[t=x^{3}-1\quad\Rightarrow\quad dt=3x^{2}\,dx,\quad x^{3}=t+1.\] Then \[\int\left(x^{3}-1\right)^{1/3}x^{5}\,dx=\frac{1}{3}\int t^{1/3}(t+1)\,dt=\frac{1}{3}\int\left(t^{4/3}+t^{1/3}\right)dt.\] Integrating, \[=\frac{1}{3}\left(\frac{3}{7}t^{7/3}+\frac{3}{4}t^{4/3}\right)+\mathrm{C}=\frac{1}{7}t^{7/3}+\frac{1}{4}t^{4/3}+\mathrm{C}.\] Hence \[\int\left(x^{3}-1\right)^{1/3}x^{5}\,dx=\frac{1}{7}\left(x^{3}-1\right)^{7/3}+\frac{1}{4}\left(x^{3}-1\right)^{4/3}+\mathrm{C}.\]

  3. Exercise 13

    x2(2+3x3)3\displaystyle \frac{x^{2}}{\left(2+3 x^{3}\right)^{3}}

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    NCERT’s answer
    \(\displaystyle -\frac{1}{18\left(2+3 x^{3}\right)^{2}}+\mathrm{C}\)
    Put \[t=2+3x^{3}\quad\Rightarrow\quad dt=9x^{2}\,dx,\ \text{ i.e. }\ x^{2}\,dx=\frac{dt}{9}.\] Then \[\int\frac{x^{2}}{\left(2+3x^{3}\right)^{3}}\,dx=\frac{1}{9}\int t^{-3}\,dt=\frac{1}{9}\cdot\frac{t^{-2}}{-2}+\mathrm{C}=-\frac{1}{18t^{2}}+\mathrm{C}.\] Hence \[\int\frac{x^{2}}{\left(2+3x^{3}\right)^{3}}\,dx=-\frac{1}{18\left(2+3x^{3}\right)^{2}}+\mathrm{C}.\]

  4. Exercise 14

    1x(log⁡x)m,x>0,m≠1\displaystyle \frac{1}{x(\log x)^{m}}, x>0, m \neq 1

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    NCERT’s answer
    \(\displaystyle \frac{(\log x)^{1-m}}{1-m}+\mathrm{C}\)
    Put \[t=\log x\quad\Rightarrow\quad dt=\frac{1}{x}\,dx\qquad(x>0).\] Then \[\int\frac{dx}{x(\log x)^{m}}=\int t^{-m}\,dt.\] The power rule \(\displaystyle \int t^{n}\,dt=\dfrac{t^{n+1}}{n+1}\) applies because \(\displaystyle -m\neq-1\), which is exactly why the question stipulates \(\displaystyle m\neq1\): \[\int t^{-m}\,dt=\frac{t^{-m+1}}{1-m}+\mathrm{C}.\] Hence \[\int\frac{1}{x(\log x)^{m}}\,dx=\frac{(\log x)^{1-m}}{1-m}+\mathrm{C},\qquad x>0,\ m\neq1.\]

  5. Exercise 15

    x9−4x2\displaystyle \frac{x}{9-4 x^{2}}

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    NCERT’s answer
    \(\displaystyle -\frac{1}{8} \log \left|9-4 \mathrm{x}^{2}\right|+\mathrm{C}\)
    The numerator is a constant multiple of the derivative of the denominator. Put \[t=9-4x^{2}\quad\Rightarrow\quad dt=-8x\,dx,\ \text{ i.e. }\ x\,dx=-\frac{dt}{8}.\] Then \[\int\frac{x}{9-4x^{2}}\,dx=-\frac{1}{8}\int\frac{dt}{t}=-\frac{1}{8}\log|t|+\mathrm{C}.\] Since \(\displaystyle 9-4x^{2}\) can be negative, the modulus must be kept: \[\int\frac{x}{9-4x^{2}}\,dx=-\frac{1}{8}\log\left|9-4x^{2}\right|+\mathrm{C}.\]

  6. Exercise 16

    e2x+3\displaystyle e^{2 x+3}

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    NCERT’s answer
    \(\displaystyle \frac{1}{2} e^{2 x+3}+\mathrm{C}\)
    Put \[t=2x+3\quad\Rightarrow\quad dt=2\,dx,\ \text{ i.e. }\ dx=\frac{dt}{2}.\] Using \(\displaystyle \int e^{t}\,dt=e^{t}+\mathrm{C}\), \[\int e^{2x+3}\,dx=\frac{1}{2}\int e^{t}\,dt=\frac{1}{2}e^{t}+\mathrm{C}.\] Hence \[\int e^{2x+3}\,dx=\frac{1}{2}e^{2x+3}+\mathrm{C}.\]

  7. Exercise 17

    xex2\displaystyle \frac{x}{e^{x^{2}}}

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    NCERT’s answer
    \(\displaystyle -\frac{1}{2 e^{x^{2}}}+\mathrm{C}\)
    Write the integrand as \(\displaystyle \dfrac{x}{e^{x^{2}}}=x\,e^{-x^{2}}\) and put \[t=x^{2}\quad\Rightarrow\quad dt=2x\,dx,\ \text{ i.e. }\ x\,dx=\frac{dt}{2}.\] Then \[\int x\,e^{-x^{2}}\,dx=\frac{1}{2}\int e^{-t}\,dt=\frac{1}{2}\left(-e^{-t}\right)+\mathrm{C}=-\frac{1}{2}e^{-t}+\mathrm{C}.\] Hence \[\int\frac{x}{e^{x^{2}}}\,dx=-\frac{1}{2e^{x^{2}}}+\mathrm{C}.\]

  8. Exercise 18

    etan⁡−1x1+x2\displaystyle \frac{e^{\tan ^{-1} x}}{1+x^{2}}

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    NCERT’s answer
    \(\displaystyle e^{\tan ^{-1} x}+\mathrm{C}\)
    Recall \(\displaystyle \dfrac{d}{dx}\left(\tan^{-1}x\right)=\dfrac{1}{1+x^{2}}\), so the factor \(\displaystyle \dfrac{dx}{1+x^{2}}\) is exactly the differential of the exponent. Put \[t=\tan^{-1}x\quad\Rightarrow\quad dt=\frac{dx}{1+x^{2}}.\] Then \[\int\frac{e^{\tan^{-1}x}}{1+x^{2}}\,dx=\int e^{t}\,dt=e^{t}+\mathrm{C}.\] Hence \[\int\frac{e^{\tan^{-1}x}}{1+x^{2}}\,dx=e^{\tan^{-1}x}+\mathrm{C}.\]

  9. Exercise 19

    e2x−1e2x+1\displaystyle \frac{e^{2 x}-1}{e^{2 x}+1}

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    NCERT’s answer
    \(\displaystyle \log \left(e^{x}+e^{-x}\right)+\mathrm{C}\)
    Divide numerator and denominator by \(\displaystyle e^{x}\) — this is the step that turns the integrand into a derivative-over-function form: \[\frac{e^{2x}-1}{e^{2x}+1}=\frac{e^{x}-e^{-x}}{e^{x}+e^{-x}}.\] Put \[t=e^{x}+e^{-x}\quad\Rightarrow\quad dt=\left(e^{x}-e^{-x}\right)dx.\] Then \[\int\frac{e^{2x}-1}{e^{2x}+1}\,dx=\int\frac{dt}{t}=\log|t|+\mathrm{C}.\] Since \(\displaystyle e^{x}+e^{-x}>0\) always, \[\int\frac{e^{2x}-1}{e^{2x}+1}\,dx=\log\left(e^{x}+e^{-x}\right)+\mathrm{C}.\]

  10. Exercise 20

    e2x−e−2xe2x+e−2x\displaystyle \frac{e^{2 x}-e^{-2 x}}{e^{2 x}+e^{-2 x}}

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    NCERT’s answer
    \(\displaystyle \frac{1}{2} \log \left(e^{2 x}+e^{-2 x}\right)+\mathrm{C}\)
    The numerator is, up to a factor \(\displaystyle 2\), the derivative of the denominator: \[\frac{d}{dx}\left(e^{2x}+e^{-2x}\right)=2\left(e^{2x}-e^{-2x}\right).\] Put \[t=e^{2x}+e^{-2x}\quad\Rightarrow\quad dt=2\left(e^{2x}-e^{-2x}\right)dx.\] Then \[\int\frac{e^{2x}-e^{-2x}}{e^{2x}+e^{-2x}}\,dx=\frac{1}{2}\int\frac{dt}{t}=\frac{1}{2}\log|t|+\mathrm{C}.\] Since \(\displaystyle e^{2x}+e^{-2x}>0\), \[\int\frac{e^{2x}-e^{-2x}}{e^{2x}+e^{-2x}}\,dx=\frac{1}{2}\log\left(e^{2x}+e^{-2x}\right)+\mathrm{C}.\]