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# Evaluate the following integral:

Question:

Evaluate the following integral:

$\int \frac{x^{2}}{\left(x^{2}+a^{2}\right)\left(x^{2}+b^{2}\right)} d x$

Solution:

Let, $I=\int \frac{x^{2}}{\left(x^{2}+a^{2}\right)\left(x^{2}+b^{2}\right)} d x$

Let $x^{2}=y$

Thus, $\frac{\mathrm{x}^{2}}{\left(\mathrm{x}^{2}+\mathrm{a}^{2}\right)\left(\mathrm{x}^{2}+\mathrm{b}^{2}\right)}=\frac{\mathrm{y}}{\left(\mathrm{y}+\mathrm{a}^{2}\right)\left(\mathrm{y}+\mathrm{b}^{2}\right)}$

Now, let $\frac{y}{\left(y+a^{2}\right)\left(y+b^{2}\right)}=\frac{A}{y+a^{2}}+\frac{B}{y+b^{2}}$

$\Rightarrow y=A\left(y+b^{2}\right)+B\left(y+a^{2}\right)$

$\Rightarrow y=y(A+B)+\left(A b^{2}+B a^{2}\right)$

Equating the coefficients, we get,

$A+B=1$, and $A b^{2}+B a^{2}=0$

On solving we get, $A=-\frac{a^{2}}{b^{2}-a^{2}}$ $B=\frac{b^{2}}{b^{2}-a^{2}}$

$\therefore I=-\frac{a^{2}}{b^{2}-a^{2}} \int \frac{d x}{x^{2}+a^{2}}+\frac{b^{2}}{b^{2}-a^{2}} \int \frac{d x}{x^{2}+b^{2}}$

$\Rightarrow I=\frac{b}{b^{2}-a^{2}} \tan ^{-1}\left(\frac{x}{b}\right)-\frac{a}{b^{2}-a^{2}} \tan ^{-1}\left(\frac{x}{a}\right)+c$

Thus, $\int \frac{\mathrm{x}^{2}}{\left(\mathrm{x}^{2}+\mathrm{a}^{2}\right)\left(\mathrm{x}^{2}+\mathrm{b}^{2}\right)} \mathrm{dx}=\frac{\mathrm{b}}{\mathrm{b}^{2}-\mathrm{a}^{2}} \tan ^{-1}\left(\frac{\mathrm{x}}{\mathrm{b}}\right)-\frac{\mathrm{a}}{\mathrm{b}^{2}-\mathrm{a}^{2}} \tan ^{-1}\left(\frac{\mathrm{x}}{\mathrm{a}}\right)+\mathrm{c}$ght)+\mathrm{c}\$