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STD 12 - 7.1 inderfinite integral — JEE STD 11 Science — Question

Gujarat BoardEnglish MediumSTD 11 ScienceJEESTD 12 - 7.1 inderfinite integral4 Marks
MCQ
$\int {\frac{{{x^2}}}{{\left( {{x^2} + 1} \right)\left( {{x^2} + 4} \right)}}\,} dx$ is equal to
  • A
    $ - {\tan ^{ - 1}}x + \frac{1}{3}{\tan ^{ - 1}}\frac{x}{2} + C$
  • ✓
    $- \frac{1}{3}{\tan ^{ - 1}}x + \frac{2}{3}{\tan ^{ - 1}}\frac{x}{2} + C$
  • C
    ${\tan ^{ - 1}}x + \frac{2}{3}{\tan ^{ - 1}}\frac{x}{2} + C$
  • D
    $\frac{1}{3}{\tan ^{ - 1}}x - \frac{2}{3}{\tan ^{ - 1}}\frac{x}{2} + C$
✓

Answer

Correct option: B.
$- \frac{1}{3}{\tan ^{ - 1}}x + \frac{2}{3}{\tan ^{ - 1}}\frac{x}{2} + C$
b
consider $\frac{x^{2}}{\left(x^{2}+1\right)\left(x^{2}+4\right)}$ and put $x^{2}=y$

Then $\frac{x^{2}}{\left(x^{2}+1\right)\left(x^{2}+4\right)}=\frac{y}{(y+1)(y+4)}$

Write $\frac{y}{(y+1)(y+4)}=\frac{A}{(y+1)}+\frac{B}{(y+4)}$

so that $y=A(y+4)+B(y+1)$

$\therefore \quad A=-\frac{1}{3}  $ and $ B=\frac{4}{3}$

$\therefore \mathrm{I}=\int \frac{-\frac{1}{3}}{\left(\mathrm{x}^{2}+1\right)} \mathrm{dx}+\int \frac{4 / 3}{\left(\mathrm{x}^{2}+4\right)} \mathrm{dx}$

$=-\frac{1}{3} \tan ^{-1} \mathrm{x}+\frac{2}{3} \tan ^{-1}\left(\frac{\mathrm{x}}{2}\right)+\mathrm{C}$

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