If int log left(a^2+x^2right) d x=h(x)+C,then | vedclass

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(C) Let $I = \int \log \left(a^2+x^2\right) d x$. Using integration by parts,$\int u v dx = u \int v dx - \int (u' \int v dx) dx$. Let $u = \log(a^2+x

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$x \log \left(a^2+x^2ight)-2 x+2 a 	an ^{-1}\left(\frac{x}{a}ight)$

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(C) Let $I = \int \log \left(a^2+x^2ight) d x$. Using integration by parts,$\int u v dx = u \int v dx - \int (u' \int v dx) dx$. Let $u = \log(a^2+x^2)$ and $v = 1$. Then $u' = \frac{2x}{a^2+x^2}$ and $\int v dx = x$. $I = x \log(a^2+x^2) - \int \frac{2x^2}{a^2+x^2} dx + C$. $I = x \log(a^2+x^2) - 2 \int \frac{x^2+a^2-a^2}{a^2+x^2} dx + C$. $I = x \log(a^2+x^2) - 2 \int (1 - \frac{a^2}{a^2+x^2}) dx + C$. $I = x \log(a^2+x^2) - 2x + 2a^2 \int \frac{1}{a^2+x^2} dx + C$. Since $\int \frac{1}{a^2+x^2} dx = \frac{1}{a} 	an^{-1}(\frac{x}{a})$,we get: $I = x \log(a^2+x^2) - 2x + 2a^2 (\frac{1}{a} 	an^{-1}(\frac{x}{a})) + C$. $I = x \log(a^2+x^2) - 2x + 2a 	an^{-1}(\frac{x}{a}) + C$. Comparing with $h(x)+C$,we have $h(x) = x \log(a^2+x^2) - 2x + 2a 	an^{-1}(\frac{x}{a})$.

If $\int \log \left(a^2+x^2\right) d x=h(x)+C$,then $h(x)$ is equal to

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