If $f(x)=\int \frac{d x}{\left(x^2+2\right)}$ and $f(\sqrt{2})=0$, then $f(0)=$

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Aicharya · Accepted Answer

$\quad f(x)=\int \frac{d x}{x^2+(\sqrt{2})^2}=\frac{1}{\sqrt{2}} 	an ^{-1}\left(\frac{x}{\sqrt{2}}ight)+c$ Since at $x=\sqrt{2}, f(x)=0$ $ \Rightarrow 0=\frac{1}{\sqrt{2}} 	an ^{-1}(1)+c \Rightarrow c=-\frac{\pi}{4 \sqrt{2}} $ Hence $ f(0)=\frac{1}{\sqrt{2}} 	an ^{-1}\left(\frac{0}{\sqrt{2}}ight)+\left(-\frac{\pi}{4 \sqrt{2}}ight)=-\frac{\pi}{4 \sqrt{2}} $

If $f(x)=\int \frac{d x}{\left(x^2+2\right)}$ and $f(\sqrt{2})=0$, then $f(0)=$

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