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NCERT Solutions · Class 12 Mathematics Integrals

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EXERCISE 7.5 11–23 (part 14 of 27)

  1. Integrate the rational functions in Exercises $\displaystyle 1$ to 21.

    Exercise 11

    5x(x+1)(x2−4)\displaystyle \frac{5 x}{(x+1)\left(x^{2}-4\right)}

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    NCERT’s answer
    \(\displaystyle \frac{5}{3} \log |x+1|-\frac{5}{2} \log |x+2|+\frac{5}{6} \log |x-2|+\mathrm{C}\)
    Since \(\displaystyle x^{2}-4=(x-2)(x+2) \), there are three distinct linear factors: \[\frac{5x}{(x+1)(x-2)(x+2)}=\frac{A}{x+1}+\frac{B}{x-2}+\frac{C}{x+2} \] \[5x=A(x-2)(x+2)+B(x+1)(x+2)+C(x+1)(x-2) \] Put \(\displaystyle x=-1 \): \(\displaystyle -5=A(-3)(1)=-3A\Rightarrow A=\tfrac53 \). Put \(\displaystyle x=2 \): \(\displaystyle 10=B(3)(4)=12B\Rightarrow B=\tfrac56 \). Put \(\displaystyle x=-2 \): \(\displaystyle -10=C(-1)(-4)=4C\Rightarrow C=-\tfrac52 \).\[\int\frac{5x\,dx}{(x+1)\left(x^{2}-4\right)}=\frac53\log|x+1|+\frac56\log|x-2|-\frac52\log|x+2|+\mathrm{C} \]

  2. Exercise 12

    x3+x+1x2−1\displaystyle \frac{x^{3}+x+1}{x^{2}-1}

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    NCERT’s answer
    \(\displaystyle \frac{x^{2}}{2}+\frac{1}{2} \log |x+1|+\frac{3}{2} \log |x-1|+C\)
    The integrand is improper (degree $\displaystyle 3$ over degree $\displaystyle 2$), so divide first. Since \(\displaystyle x^{3}+x+1=x\left(x^{2}-1\right)+2x+1 \), \[\frac{x^{3}+x+1}{x^{2}-1}=x+\frac{2x+1}{x^{2}-1} \] Now split the proper part: \[\frac{2x+1}{(x-1)(x+1)}=\frac{A}{x-1}+\frac{B}{x+1},\qquad 2x+1=A(x+1)+B(x-1) \] Put \(\displaystyle x=1 \): \(\displaystyle 3=2A\Rightarrow A=\tfrac32 \). Put \(\displaystyle x=-1 \): \(\displaystyle -1=-2B\Rightarrow B=\tfrac12 \).\[\int\frac{x^{3}+x+1}{x^{2}-1}\,dx=\frac{x^{2}}{2}+\frac32\log|x-1|+\frac12\log|x+1|+\mathrm{C} \]

  3. Exercise 13

    2(1−x)(1+x2)\displaystyle \frac{2}{(1-x)\left(1+x^{2}\right)}

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    NCERT’s answer
    \(\displaystyle -\log |x-1|+\frac{1}{2} \log \left(1+x^{2}\right)+\tan ^{-1} x+\mathrm{C}\)
    \(\displaystyle 1+x^{2} \) is irreducible, so it takes a linear numerator: \[\frac{2}{(1-x)\left(1+x^{2}\right)}=\frac{A}{1-x}+\frac{Bx+C}{1+x^{2}} \] \[2=A\left(1+x^{2}\right)+(Bx+C)(1-x) \] Put \(\displaystyle x=1 \): \(\displaystyle 2=2A\Rightarrow A=1 \). Coefficient of \(\displaystyle x^{2} \): \(\displaystyle 0=A-B\Rightarrow B=1 \). Constant term: \(\displaystyle 2=A+C\Rightarrow C=1 \).So the integrand is \(\displaystyle \dfrac{1}{1-x}+\dfrac{x}{1+x^{2}}+\dfrac{1}{1+x^{2}} \). Note \(\displaystyle \displaystyle\int\frac{dx}{1-x}=-\log|1-x| \) (the inner derivative is \(\displaystyle -1 \)). \[\int\frac{2\,dx}{(1-x)\left(1+x^{2}\right)}=-\log|1-x|+\frac12\log\left(1+x^{2}\right)+\tan^{-1}x+\mathrm{C} \]

  4. Exercise 14

    3x−1(x+2)2\displaystyle \frac{3 x-1}{(x+2)^{2}}

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    NCERT’s answer
    \(\displaystyle 3 \log |x+2|+\frac{7}{x+2}+C\)
    A single repeated linear factor; substitute \(\displaystyle t=x+2 \), so \(\displaystyle x=t-2 \) and \(\displaystyle dx=dt \): \[\frac{3x-1}{(x+2)^{2}}=\frac{3(t-2)-1}{t^{2}}=\frac{3t-7}{t^{2}}=\frac{3}{t}-\frac{7}{t^{2}} \] \[\int\left(\frac{3}{t}-\frac{7}{t^{2}}\right)dt=3\log|t|+\frac{7}{t}+\mathrm{C} \] Restoring \(\displaystyle t=x+2 \): \[\int\frac{(3x-1)\,dx}{(x+2)^{2}}=3\log|x+2|+\frac{7}{x+2}+\mathrm{C} \]

  5. Exercise 15

    1x4−1\displaystyle \frac{1}{x^{4}-1}

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    NCERT’s answer
    \(\displaystyle \frac{1}{4} \log \left|\frac{x-1}{x+1}\right|-\frac{1}{2} \tan ^{-1} x+\mathrm{C}\)
    Factorise: \(\displaystyle x^{4}-1=\left(x^{2}-1\right)\left(x^{2}+1\right) \). Split on these two blocks first, using \(\displaystyle \left(x^{2}+1\right)-\left(x^{2}-1\right)=2 \): \[\frac{1}{\left(x^{2}-1\right)\left(x^{2}+1\right)}=\frac12\left(\frac{1}{x^{2}-1}-\frac{1}{x^{2}+1}\right) \] The first piece splits further, since \(\displaystyle x^{2}-1=(x-1)(x+1) \): \[\frac{1}{x^{2}-1}=\frac12\left(\frac{1}{x-1}-\frac{1}{x+1}\right) \] Therefore \[\int\frac{dx}{x^{4}-1}=\frac14\int\left(\frac{1}{x-1}-\frac{1}{x+1}\right)dx-\frac12\int\frac{dx}{x^{2}+1} \] \[=\frac14\log\left|\frac{x-1}{x+1}\right|-\frac12\tan^{-1}x+\mathrm{C} \]

  6. Exercise 16

    1x(xn+1)\displaystyle \frac{1}{x\left(x^{n}+1\right)} [Hint: multiply numerator and denominator by xn−1\displaystyle x^{n-1} and put xn=t\displaystyle x^{n}=t ]

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    NCERT’s answer
    \(\displaystyle \frac{1}{n} \log \left|\frac{x^{n}}{x^{n}+1}\right|+\mathrm{C}\)
    Following the hint, multiply numerator and denominator by \(\displaystyle x^{n-1} \): \[\frac{1}{x\left(x^{n}+1\right)}=\frac{x^{n-1}}{x^{n}\left(x^{n}+1\right)} \] Put \(\displaystyle t=x^{n} \), so \(\displaystyle dt=n\,x^{n-1}dx \), i.e. \(\displaystyle x^{n-1}dx=\dfrac{dt}{n} \): \[\int\frac{dx}{x\left(x^{n}+1\right)}=\frac1n\int\frac{dt}{t(t+1)} \] Since \(\displaystyle \dfrac{1}{t(t+1)}=\dfrac1t-\dfrac{1}{t+1} \), \[=\frac1n\left(\log|t|-\log|t+1|\right)+\mathrm{C}=\frac1n\log\left|\frac{t}{t+1}\right|+\mathrm{C} \] Restoring \(\displaystyle t=x^{n} \): \[\int\frac{dx}{x\left(x^{n}+1\right)}=\frac1n\log\left|\frac{x^{n}}{x^{n}+1}\right|+\mathrm{C} \]

  7. Exercise 17

    cos⁡x(1−sin⁡x)(2−sin⁡x)\displaystyle \frac{\cos x}{(1-\sin x)(2-\sin x)} [Hint : Put sin⁡x=t\displaystyle \sin x=t ]

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    NCERT’s answer
    \(\displaystyle \log \left|\frac{2-\sin x}{1-\sin x}\right|+C\)
    Put \(\displaystyle t=\sin x \), so \(\displaystyle dt=\cos x\,dx \) — the numerator is exactly \(\displaystyle dt \): \[\int\frac{\cos x\,dx}{(1-\sin x)(2-\sin x)}=\int\frac{dt}{(1-t)(2-t)} \] Partial fractions: \[\frac{1}{(1-t)(2-t)}=\frac{A}{1-t}+\frac{B}{2-t},\qquad 1=A(2-t)+B(1-t) \] Put \(\displaystyle t=1 \): \(\displaystyle 1=A \). Put \(\displaystyle t=2 \): \(\displaystyle 1=-B\Rightarrow B=-1 \).Watch the signs: \(\displaystyle \displaystyle\int\frac{dt}{1-t}=-\log|1-t| \) and \(\displaystyle \displaystyle\int\frac{dt}{2-t}=-\log|2-t| \). \[\int\left(\frac{1}{1-t}-\frac{1}{2-t}\right)dt=-\log|1-t|+\log|2-t|+\mathrm{C} \] Restoring \(\displaystyle t=\sin x \): \[=\log\left|\frac{2-\sin x}{1-\sin x}\right|+\mathrm{C} \]

  8. Exercise 18

    (x2+1)(x2+2)(x2+3)(x2+4)\displaystyle \frac{\left(x^{2}+1\right)\left(x^{2}+2\right)}{\left(x^{2}+3\right)\left(x^{2}+4\right)}

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    NCERT’s answer
    \(\displaystyle x+\frac{2}{\sqrt{3}} \tan ^{-1} \frac{x}{\sqrt{3}}-3 \tan ^{-1} \frac{x}{2}+C\)
    Numerator and denominator both have degree $\displaystyle 4$, so make the fraction proper first: \[\frac{\left(x^{2}+1\right)\left(x^{2}+2\right)}{\left(x^{2}+3\right)\left(x^{2}+4\right)}=\frac{x^{4}+3x^{2}+2}{x^{4}+7x^{2}+12}=1-\frac{4x^{2}+10}{\left(x^{2}+3\right)\left(x^{2}+4\right)} \] Now split the proper part, treating \(\displaystyle y=x^{2} \) as the variable (do not try to split into linear factors of \(\displaystyle x \) — \(\displaystyle x^{2}+3 \) and \(\displaystyle x^{2}+4 \) are irreducible): \[\frac{4y+10}{(y+3)(y+4)}=\frac{A}{y+3}+\frac{B}{y+4},\qquad 4y+10=A(y+4)+B(y+3) \] Put \(\displaystyle y=-3 \): \(\displaystyle -2=A \). Put \(\displaystyle y=-4 \): \(\displaystyle -6=-B\Rightarrow B=6 \).Hence the integrand equals \[1-\left(\frac{-2}{x^{2}+3}+\frac{6}{x^{2}+4}\right)=1+\frac{2}{x^{2}+3}-\frac{6}{x^{2}+4} \] Using \(\displaystyle \displaystyle\int\frac{dx}{x^{2}+a^{2}}=\frac1a\tan^{-1}\frac{x}{a} \) with \(\displaystyle a=\sqrt3 \) and \(\displaystyle a=2 \): \[\int\frac{\left(x^{2}+1\right)\left(x^{2}+2\right)}{\left(x^{2}+3\right)\left(x^{2}+4\right)}dx=x+\frac{2}{\sqrt3}\tan^{-1}\frac{x}{\sqrt3}-3\tan^{-1}\frac{x}{2}+\mathrm{C} \]

  9. Exercise 19

    2x(x2+1)(x2+3)\displaystyle \frac{2 x}{\left(x^{2}+1\right)\left(x^{2}+3\right)}

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    NCERT’s answer
    \(\displaystyle \frac{1}{2} \log \left(\frac{x^{2}+1}{x^{2}+3}\right)+\mathrm{C}\)
    The numerator \(\displaystyle 2x\,dx \) is the differential of \(\displaystyle x^{2} \), so put \(\displaystyle t=x^{2} \), \(\displaystyle dt=2x\,dx \): \[\int\frac{2x\,dx}{\left(x^{2}+1\right)\left(x^{2}+3\right)}=\int\frac{dt}{(t+1)(t+3)} \] Partial fractions in \(\displaystyle t \): \[\frac{1}{(t+1)(t+3)}=\frac12\left(\frac{1}{t+1}-\frac{1}{t+3}\right) \] \[\int\frac{dt}{(t+1)(t+3)}=\frac12\log\left|\frac{t+1}{t+3}\right|+\mathrm{C} \] Restoring \(\displaystyle t=x^{2} \) (both \(\displaystyle x^{2}+1 \) and \(\displaystyle x^{2}+3 \) are positive, so the modulus may be dropped): \[=\frac12\log\frac{x^{2}+1}{x^{2}+3}+\mathrm{C} \]

  10. Exercise 20

    1x(x4−1)\displaystyle \frac{1}{x\left(x^{4}-1\right)}

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    NCERT’s answer
    \(\displaystyle \frac{1}{4} \log \left|\frac{x^{4}-1}{x^{4}}\right|+\mathrm{C}\)
    Multiply numerator and denominator by \(\displaystyle x^{3} \) so that the substitution \(\displaystyle t=x^{4} \) becomes available: \[\frac{1}{x\left(x^{4}-1\right)}=\frac{x^{3}}{x^{4}\left(x^{4}-1\right)} \] Put \(\displaystyle t=x^{4} \), so \(\displaystyle dt=4x^{3}dx \), i.e. \(\displaystyle x^{3}dx=\dfrac{dt}{4} \): \[\int\frac{dx}{x\left(x^{4}-1\right)}=\frac14\int\frac{dt}{t(t-1)} \] Since \(\displaystyle \dfrac{1}{t(t-1)}=\dfrac{1}{t-1}-\dfrac{1}{t} \), \[=\frac14\left(\log|t-1|-\log|t|\right)+\mathrm{C}=\frac14\log\left|\frac{t-1}{t}\right|+\mathrm{C} \] Restoring \(\displaystyle t=x^{4} \): \[\int\frac{dx}{x\left(x^{4}-1\right)}=\frac14\log\left|\frac{x^{4}-1}{x^{4}}\right|+\mathrm{C} \]

  11. Exercise 21

    1(ex−1)\displaystyle \frac{1}{\left(e^{x}-1\right)} [Hint : Put ex=t\displaystyle e^{x}=t ]

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    NCERT’s answer
    \(\displaystyle \log \left(\frac{e^{x}-1}{e^{x}}\right)+\mathrm{C}\)
    Following the hint, put \(\displaystyle t=e^{x} \). Then \(\displaystyle dt=e^{x}dx=t\,dx \), so \(\displaystyle dx=\dfrac{dt}{t} \) — this extra \(\displaystyle \frac1t \) is the step most often dropped. \[\int\frac{dx}{e^{x}-1}=\int\frac{1}{t-1}\cdot\frac{dt}{t}=\int\frac{dt}{t(t-1)} \] Partial fractions: \(\displaystyle \dfrac{1}{t(t-1)}=\dfrac{1}{t-1}-\dfrac{1}{t} \). \[\int\left(\frac{1}{t-1}-\frac{1}{t}\right)dt=\log|t-1|-\log|t|+\mathrm{C} \] Restoring \(\displaystyle t=e^{x} \) and using \(\displaystyle \log e^{x}=x \): \[\int\frac{dx}{e^{x}-1}=\log\left|e^{x}-1\right|-x+\mathrm{C} \]
  12. Choose the correct answer in each of the Exercises $\displaystyle 22$ and 23.

    Exercise 22

    ∫xdx(x−1)(x−2)\displaystyle \int \frac{x d x}{(x-1)(x-2)} equals (A) log⁡∣(x−1)2x−2∣+C\displaystyle \log \left|\frac{(x-1)^{2}}{x-2}\right|+\mathrm{C} (B) log⁡∣(x−2)2x−1∣+C\displaystyle \log \left|\frac{(x-2)^{2}}{x-1}\right|+\mathrm{C} (C) log⁡∣(x−1x−2)2∣+C\displaystyle \log \left|\left(\frac{x-1}{x-2}\right)^{2}\right|+\mathrm{C} (D) log⁡∣(x−1)(x−2)∣+C\displaystyle \log |(x-1)(x-2)|+\mathrm{C}

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    NCERT’s answer
    B
    Partial fractions with two distinct linear factors: \[\frac{x}{(x-1)(x-2)}=\frac{A}{x-1}+\frac{B}{x-2},\qquad x=A(x-2)+B(x-1) \] Put \(\displaystyle x=1 \): \(\displaystyle 1=-A\Rightarrow A=-1 \). Put \(\displaystyle x=2 \): \(\displaystyle 2=B \).\[\int\frac{x\,dx}{(x-1)(x-2)}=-\log|x-1|+2\log|x-2|+\mathrm{C}=\log\left|\frac{(x-2)^{2}}{x-1}\right|+\mathrm{C} \]The correct answer is (B).

  13. Exercise 23

    ∫dxx(x2+1)\displaystyle \int \frac{d x}{x\left(x^{2}+1\right)} equals (A) log⁡∣x∣−12log⁡(x2+1)+C\displaystyle \log |x|-\frac{1}{2} \log \left(x^{2}+1\right)+C (B) log⁡∣x∣+12log⁡(x2+1)+C\displaystyle \log |x|+\frac{1}{2} \log \left(x^{2}+1\right)+C (C) −log⁡∣x∣+12log⁡(x2+1)+C\displaystyle -\log |x|+\frac{1}{2} \log \left(x^{2}+1\right)+C (D) 12log⁡∣x∣+log⁡(x2+1)+C\displaystyle \frac{1}{2} \log |x|+\log \left(x^{2}+1\right)+C

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    NCERT’s answer
    A
    \(\displaystyle x^{2}+1 \) is irreducible, so it takes a linear numerator: \[\frac{1}{x\left(x^{2}+1\right)}=\frac{A}{x}+\frac{Bx+C}{x^{2}+1},\qquad 1=A\left(x^{2}+1\right)+(Bx+C)x \] Put \(\displaystyle x=0 \): \(\displaystyle A=1 \). Coefficient of \(\displaystyle x^{2} \): \(\displaystyle 0=A+B\Rightarrow B=-1 \). Coefficient of \(\displaystyle x \): \(\displaystyle C=0 \).\[\int\frac{dx}{x\left(x^{2}+1\right)}=\int\frac{dx}{x}-\int\frac{x\,dx}{x^{2}+1}=\log|x|-\frac12\log\left(x^{2}+1\right)+\mathrm{C} \]The correct answer is (A).