$\int \frac{5 x^2+3}{x^2\left(x^2-2\right)} d x=$

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Aicharya · Accepted Answer

Let, $I=\int \frac{5 x^2+3}{x^2\left(x^2-2\right)} d x$ Put $x^2=y$. Then,

$ \begin{aligned} & 5 y+3=(y-2) A+y B \\ & 5 y+3=y(A+B)-2 A \end{aligned} $ On comparing the coefficients, we get $A+B=5$ and $3=-2 A$ $A=\frac{-3}{2}$ and $B=\frac{13}{2}$ Thus, $\frac{5 y+3}{y(y-2)}=\frac{5 x^2+3}{x^2\left(x^2-2\right)}=-\frac{3}{2} \frac{1}{x^2}+\frac{13}{2} \frac{1}{x^2-2}$ Now, $I=-\frac{3}{2} \int \frac{d x}{x^2}+\frac{13}{2} \int \frac{d x}{x^2-2}$ $ \begin{gathered} =-\frac{3}{2}\left(-\frac{1}{x}\right)+\frac{13}{2} \times \frac{1}{2 \sqrt{2}} \log \left|\frac{x-\sqrt{2}}{x+\sqrt{2}}\right|+C \\ =\frac{3}{2 x}+\frac{\sqrt{13}}{4 \sqrt{2}} \log \left|\frac{x-\sqrt{2}}{x+\sqrt{2}}\right|+C \end{gathered} $

$\int \frac{5 x^2+3}{x^2\left(x^2-2\right)} d x=$

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