Gujarat BoardEnglish MediumSTD 12 ScienceMathsModel Paper 13 Marks
Question
Evaluate: $\int \frac{x^2+1}{\left(x^2+2\right)\left(2 x^2+1\right)} d x$
✓
Answer
Let $x^2=y$. Then,
$\frac{x^2+1}{\left(x^2+2\right)\left(2 x^2+1\right)}=\frac{y+1}{(y+2)(2 y+1)} $
Using partial fractions,
$\frac{y+1}{(y+2)(2 y+1)}=\frac{A}{y+2}+\frac{B}{2 y+1} \ldots \ldots \text { (i) }$
$\Rightarrow y+1=A(2 y+1)+B(y+2)$
Putting $y+2=0$
i.e. $y=-2$ in $(ii),$ we get
$-1=-3 A \Rightarrow A=\frac{1}{3}$
Putting $2 y +1=0$ i.e. $y =-\frac{1}{2}$ in $(ii),$ we get
$\frac{1}{2}=B\left(\frac{3}{2}\right)$
$\Rightarrow B=\frac{1}{3}$
Substituting the values of $A$ and $B$ in $(i)$, we obtain
$\frac{y+1}{(y+2)(2 y+1)}=\frac{1}{3} \cdot \frac{1}{y+2}+\frac{1}{3} \frac{1}{(2 y+1)}$
Replacing $y$ by $x^2$, we get
$ \frac{x^2+1}{\left(x^2+2\right)\left(2 x^2+1\right)}=\frac{1}{3} \cdot \frac{1}{x^2+2}+\frac{1}{3\left(2 x^2+1\right)}$
$\therefore I=\int \frac{x^2+1}{\left(x^2+2\right)\left(2 x^2+1\right)} d x=\frac{1}{3} \int \frac{1}{x^2+2} d x+\frac{1}{3} \int \frac{1}{(\sqrt{2} x)^2+1} d x$
$\Rightarrow I=\frac{1}{3} \times \frac{1}{\sqrt{2}} \tan ^{-1}\left(\frac{x}{\sqrt{2}}\right)+\frac{1}{3 \sqrt{2}} \tan ^{-1}(\sqrt{2} x)+C$
$\Rightarrow I=\frac{1}{3 \sqrt{2}}\left\{\tan ^{-1} \frac{x}{\sqrt{2}}+\tan ^{-1} \sqrt{2}x\right\}+C$
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