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NCERT Solutions · Class 12 Mathematics Integrals

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EXERCISE 7.5 1–10 (part 13 of 27)

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

    Exercise 1

    x(x+1)(x+2)\displaystyle \frac{x}{(x+1)(x+2)}

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    NCERT’s answer
    \(\displaystyle \log \frac{(x+2)^{2}}{|x+1|}+C\)
    The denominator has two distinct linear factors, so use partial fractions: \[\frac{x}{(x+1)(x+2)}=\frac{A}{x+1}+\frac{B}{x+2} \] Clearing denominators, \(\displaystyle x=A(x+2)+B(x+1) \).Put \(\displaystyle x=-1 \): \(\displaystyle -1=A(1)\Rightarrow A=-1 \). Put \(\displaystyle x=-2 \): \(\displaystyle -2=B(-1)\Rightarrow B=2 \).\[\int\frac{x\,dx}{(x+1)(x+2)}=\int\left(\frac{-1}{x+1}+\frac{2}{x+2}\right)dx \] \[=-\log|x+1|+2\log|x+2|+\mathrm{C} \]

  2. Exercise 2

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

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    NCERT’s answer
    \(\displaystyle \frac{1}{6} \log \left|\frac{x-3}{x+3}\right|+\mathrm{C}\)
    Factorise: \(\displaystyle x^{2}-9=(x-3)(x+3) \), two distinct linear factors. \[\frac{1}{x^{2}-9}=\frac{A}{x-3}+\frac{B}{x+3},\qquad 1=A(x+3)+B(x-3) \] Put \(\displaystyle x=3 \): \(\displaystyle 1=6A\Rightarrow A=\tfrac16 \). Put \(\displaystyle x=-3 \): \(\displaystyle 1=-6B\Rightarrow B=-\tfrac16 \).\[\int\frac{dx}{x^{2}-9}=\frac16\int\left(\frac{1}{x-3}-\frac{1}{x+3}\right)dx=\frac16\log\left|\frac{x-3}{x+3}\right|+\mathrm{C} \]

  3. Exercise 3

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

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    NCERT’s answer
    \(\displaystyle \log |x-1|-5 \log |x-2|+4 \log |x-3|+C\)
    Three distinct linear factors: \[\frac{3x-1}{(x-1)(x-2)(x-3)}=\frac{A}{x-1}+\frac{B}{x-2}+\frac{C}{x-3} \] \[3x-1=A(x-2)(x-3)+B(x-1)(x-3)+C(x-1)(x-2) \] Put \(\displaystyle x=1 \): \(\displaystyle 2=A(-1)(-2)=2A\Rightarrow A=1 \). Put \(\displaystyle x=2 \): \(\displaystyle 5=B(1)(-1)\Rightarrow B=-5 \). Put \(\displaystyle x=3 \): \(\displaystyle 8=C(2)(1)\Rightarrow C=4 \).\[\int\frac{(3x-1)\,dx}{(x-1)(x-2)(x-3)}=\log|x-1|-5\log|x-2|+4\log|x-3|+\mathrm{C} \]

  4. Exercise 4

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

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    NCERT’s answer
    \(\displaystyle \frac{1}{2} \log |x-1|-2 \log |x-2|+\frac{3}{2} \log |x-3|+\mathrm{C}\)
    Three distinct linear factors: \[\frac{x}{(x-1)(x-2)(x-3)}=\frac{A}{x-1}+\frac{B}{x-2}+\frac{C}{x-3} \] \[x=A(x-2)(x-3)+B(x-1)(x-3)+C(x-1)(x-2) \] Put \(\displaystyle x=1 \): \(\displaystyle 1=A(-1)(-2)=2A\Rightarrow A=\tfrac12 \). Put \(\displaystyle x=2 \): \(\displaystyle 2=B(1)(-1)\Rightarrow B=-2 \). Put \(\displaystyle x=3 \): \(\displaystyle 3=C(2)(1)\Rightarrow C=\tfrac32 \).\[\int\frac{x\,dx}{(x-1)(x-2)(x-3)}=\frac12\log|x-1|-2\log|x-2|+\frac32\log|x-3|+\mathrm{C} \]

  5. Exercise 5

    2xx2+3x+2\displaystyle \frac{2 x}{x^{2}+3 x+2}

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    NCERT’s answer
    \(\displaystyle 4 \log |x+2|-2 \log |x+1|+C\)
    Factorise the denominator: \(\displaystyle x^{2}+3x+2=(x+1)(x+2) \). \[\frac{2x}{(x+1)(x+2)}=\frac{A}{x+1}+\frac{B}{x+2},\qquad 2x=A(x+2)+B(x+1) \] Put \(\displaystyle x=-1 \): \(\displaystyle -2=A(1)\Rightarrow A=-2 \). Put \(\displaystyle x=-2 \): \(\displaystyle -4=B(-1)\Rightarrow B=4 \).\[\int\frac{2x\,dx}{x^{2}+3x+2}=-2\log|x+1|+4\log|x+2|+\mathrm{C} \]

  6. Exercise 6

    1−x2x(1−2x)\displaystyle \frac{1-x^{2}}{x(1-2 x)}

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    NCERT’s answer
    \(\displaystyle \frac{x}{2}+\log |x|-\frac{3}{4} \log |1-2 x|+C\)
    The integrand is an improper rational function: the numerator \(\displaystyle 1-x^{2} \) and the denominator \(\displaystyle x(1-2x)=x-2x^{2} \) both have degree $\displaystyle 2$, so divide before splitting.Since \(\displaystyle \tfrac12\left(-2x^{2}+x\right)=-x^{2}+\tfrac{x}{2} \), the remainder is \(\displaystyle 1-\tfrac{x}{2} \): \[\frac{1-x^{2}}{x(1-2x)}=\frac12+\frac{1-\dfrac{x}{2}}{x(1-2x)}=\frac12+\frac{2-x}{2\,x(1-2x)} \]Now split the proper part: \[\frac{2-x}{x(1-2x)}=\frac{A}{x}+\frac{B}{1-2x},\qquad 2-x=A(1-2x)+Bx \] Put \(\displaystyle x=0 \): \(\displaystyle A=2 \). Put \(\displaystyle x=\tfrac12 \): \(\displaystyle \tfrac32=\tfrac{B}{2}\Rightarrow B=3 \).So the integrand is \(\displaystyle \dfrac12+\dfrac1x+\dfrac{3}{2(1-2x)} \). Using \(\displaystyle \displaystyle\int\frac{dx}{1-2x}=-\frac12\log|1-2x| \), \[\int\frac{(1-x^{2})\,dx}{x(1-2x)}=\frac{x}{2}+\log|x|-\frac34\log|1-2x|+\mathrm{C} \]

  7. Exercise 7

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

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    NCERT’s answer
    \(\displaystyle \frac{1}{2} \log |x-1|-\frac{1}{4} \log \left(x^{2}+1\right)+\frac{1}{2} \tan ^{-1} x+C\)
    \(\displaystyle x^{2}+1 \) is an irreducible quadratic factor, so it takes a linear numerator: \[\frac{x}{(x^{2}+1)(x-1)}=\frac{Ax+B}{x^{2}+1}+\frac{C}{x-1} \] \[x=(Ax+B)(x-1)+C(x^{2}+1) \] Put \(\displaystyle x=1 \): \(\displaystyle 1=2C\Rightarrow C=\tfrac12 \). Coefficient of \(\displaystyle x^{2} \): \(\displaystyle 0=A+C\Rightarrow A=-\tfrac12 \). Constant term: \(\displaystyle 0=-B+C\Rightarrow B=\tfrac12 \).\[\int\frac{x\,dx}{(x^{2}+1)(x-1)}=-\frac12\int\frac{x\,dx}{x^{2}+1}+\frac12\int\frac{dx}{x^{2}+1}+\frac12\int\frac{dx}{x-1} \] \[=-\frac14\log\left(x^{2}+1\right)+\frac12\tan^{-1}x+\frac12\log|x-1|+\mathrm{C} \]

  8. Exercise 8

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

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    NCERT’s answer
    \(\displaystyle \frac{2}{9} \log \left|\frac{x-1}{x+2}\right|-\frac{1}{3(x-1)}+\mathrm{C}\)
    \(\displaystyle (x-1) \) is a repeated factor, so both powers must appear: \[\frac{x}{(x-1)^{2}(x+2)}=\frac{A}{x-1}+\frac{B}{(x-1)^{2}}+\frac{C}{x+2} \] \[x=A(x-1)(x+2)+B(x+2)+C(x-1)^{2} \] Put \(\displaystyle x=1 \): \(\displaystyle 1=3B\Rightarrow B=\tfrac13 \). Put \(\displaystyle x=-2 \): \(\displaystyle -2=9C\Rightarrow C=-\tfrac29 \). Coefficient of \(\displaystyle x^{2} \): \(\displaystyle 0=A+C\Rightarrow A=\tfrac29 \).\[\int\frac{x\,dx}{(x-1)^{2}(x+2)}=\frac29\log|x-1|+\frac13\cdot\left(-\frac{1}{x-1}\right)-\frac29\log|x+2|+\mathrm{C} \] \[=\frac29\log\left|\frac{x-1}{x+2}\right|-\frac{1}{3(x-1)}+\mathrm{C} \]

  9. Exercise 9

    3x+5x3−x2−x+1\displaystyle \frac{3 x+5}{x^{3}-x^{2}-x+1}

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    NCERT’s answer
    \(\displaystyle \frac{1}{2} \log \left|\frac{x+1}{x-1}\right|-\frac{4}{x-1}+\mathrm{C}\)
    Factorise the denominator by grouping: \[x^{3}-x^{2}-x+1=x^{2}(x-1)-(x-1)=(x-1)\left(x^{2}-1\right)=(x-1)^{2}(x+1) \] A repeated factor \(\displaystyle (x-1) \) and a simple factor \(\displaystyle (x+1) \): \[\frac{3x+5}{(x-1)^{2}(x+1)}=\frac{A}{x-1}+\frac{B}{(x-1)^{2}}+\frac{C}{x+1} \] \[3x+5=A(x-1)(x+1)+B(x+1)+C(x-1)^{2} \] Put \(\displaystyle x=1 \): \(\displaystyle 8=2B\Rightarrow B=4 \). Put \(\displaystyle x=-1 \): \(\displaystyle 2=4C\Rightarrow C=\tfrac12 \). Coefficient of \(\displaystyle x^{2} \): \(\displaystyle 0=A+C\Rightarrow A=-\tfrac12 \).\[\int\frac{(3x+5)\,dx}{x^{3}-x^{2}-x+1}=-\frac12\log|x-1|-\frac{4}{x-1}+\frac12\log|x+1|+\mathrm{C} \] \[=\frac12\log\left|\frac{x+1}{x-1}\right|-\frac{4}{x-1}+\mathrm{C} \]

  10. Exercise 10

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

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    NCERT’s answer
    \(\displaystyle \frac{5}{2} \log |x+1|-\frac{1}{10} \log |x-1|-\frac{12}{5} \log |2 x+3|+C\)
    Since \(\displaystyle x^{2}-1=(x-1)(x+1) \), there are three distinct linear factors: \[\frac{2x-3}{(x-1)(x+1)(2x+3)}=\frac{A}{x-1}+\frac{B}{x+1}+\frac{C}{2x+3} \] \[2x-3=A(x+1)(2x+3)+B(x-1)(2x+3)+C(x-1)(x+1) \] Put \(\displaystyle x=1 \): \(\displaystyle -1=A(2)(5)=10A\Rightarrow A=-\tfrac{1}{10} \). Put \(\displaystyle x=-1 \): \(\displaystyle -5=B(-2)(1)=-2B\Rightarrow B=\tfrac52 \). Put \(\displaystyle x=-\tfrac32 \) (the zero of the third factor): \(\displaystyle -6=C\left(-\tfrac52\right)\left(-\tfrac12\right)=\tfrac54C\Rightarrow C=-\tfrac{24}{5} \).Note \(\displaystyle \displaystyle\int\frac{dx}{2x+3}=\frac12\log|2x+3| \), so the third term contributes \(\displaystyle -\tfrac{24}{5}\cdot\tfrac12=-\tfrac{12}{5} \) times the log. \[\int\frac{(2x-3)\,dx}{\left(x^{2}-1\right)(2x+3)}=-\frac{1}{10}\log|x-1|+\frac52\log|x+1|-\frac{12}{5}\log|2x+3|+\mathrm{C} \]