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Question Bank [2022] — Maths STD 12 Science — Question

Maharashtra BoardEnglish MediumSTD 12 ScienceMathsQuestion Bank [2022]4 Marks
Question
$\int \frac{x^2}{\left(x^2+1\right)\left(x^2-2\right)\left(x^2+3\right)} d x$
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Answer

$\text { Let } I =\int \frac{x^2}{\left(x^2+1\right)\left(x^2-2\right)\left(x^2+3\right)} d x$
Let $\frac{x^2}{\left(x^2+1\right)\left(x^2-2\right)\left(x^2+3\right)}$
$ =\frac{A}{x^2+1}+\frac{b}{x^2-2}+\frac{c}{x^2+3}$
$\therefore x^2=A\left(x^2-2\right)\left(x^2+3\right)+B\left(x^2+1\right)\left(x^2+3\right)+C\left(x^2+1\right)\left(x^2-2\right)\ldots(i) $
Putting $x^2=2$ in (i), we get
$ 2=B \times 3 \times 5$
$\therefore B=\frac{2}{15} $
Putting $x^2=-3$ in (i), we get
$ -3=C \times(-2) \times(-5)$
$\therefore C=\frac{-3}{10} $
Putting $x^2=-1$ in (i), we get
$ -1=A \times(-3) \times 2$
$\therefore A=\frac{1}{6} $
$\therefore \frac{x^2}{\left(x^2+1\right)\left(x^2-2\right)\left(x^2+3\right)}=\frac{\frac{1}{6}}{x^2+1}+\frac{\frac{2}{15}}{x^2-2}+\frac{\frac{-3}{10}}{x^2+3}$
$\therefore I =\int\left[\frac{1}{6\left(x^2+1\right)}+\frac{2}{15\left(x^2-2\right)}-\frac{3}{10\left(x^2+3\right)}\right] d x$
$=\frac{1}{6} \int \frac{1}{x^2+1} d x+\frac{2}{15} \int \frac{1}{x^2-2} d x-\frac{3}{10} \int \frac{1}{x^2+3} d x$
$=\frac{1}{6} \int \frac{1}{x^2+1} d x+\frac{2}{15} \int \frac{1}{x^2-(\sqrt{2})^2} d x-\frac{3}{10} \int \frac{1}{x^2+(\sqrt{3})^2} d x$
$=\frac{1}{6} \tan ^{-1} x+\frac{2}{15} \times \frac{1}{2 \times \sqrt{2}} \log \left|\frac{x-\sqrt{2}}{x+\sqrt{2}}\right|-\frac{3}{10} \times \frac{1}{\sqrt{3}} \tan ^{-1}\left(\frac{x}{\sqrt{3}}\right)+ c$
$\therefore I =\frac{1}{6} \tan ^{-1} x+\frac{1}{15 \sqrt{2}} \log \left|\frac{x-\sqrt{2}}{x+\sqrt{2}}\right|-\frac{\sqrt{3}}{10} \tan ^{-1}\left(\frac{x}{\sqrt{3}}\right)+ c $

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