If int(1+x) log left(1+x^2right) d x=left(x+f | vedclass

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(A) We are given the integral $I = \int(1+x) \log(1+x^2) dx = \int \log(1+x^2) dx + \int x \log(1+x^2) dx$.First,evaluate $I_1 = \int \log(1+x^2) dx$

Vedclass · Accepted Answer

$-2 x-\frac{x^2}{2}+2 	an ^{-1} x$

Vedclass · Answer

(A) We are given the integral $I = \int(1+x) \log(1+x^2) dx = \int \log(1+x^2) dx + \int x \log(1+x^2) dx$.First,evaluate $I_1 = \int \log(1+x^2) dx$ using integration by parts:$I_1 = x \log(1+x^2) - \int x \cdot \frac{2x}{1+x^2} dx = x \log(1+x^2) - 2 \int \frac{x^2+1-1}{1+x^2} dx$$I_1 = x \log(1+x^2) - 2 \int (1 - \frac{1}{1+x^2}) dx = x \log(1+x^2) - 2x + 2 	an^{-1} x$.Next,evaluate $I_2 = \int x \log(1+x^2) dx$. Let $1+x^2 = t$,so $2x dx = dt$ or $x dx = \frac{1}{2} dt$:$I_2 = \frac{1}{2} \int \log t dt = \frac{1}{2} (t \log t - t) = \frac{1}{2} ((1+x^2) \log(1+x^2) - (1+x^2))$.Adding $I_1$ and $I_2$:$I = x \log(1+x^2) - 2x + 2 	an^{-1} x + \frac{1}{2} (1+x^2) \log(1+x^2) - \frac{1}{2} (1+x^2)$$I = (x + \frac{1}{2} + \frac{x^2}{2}) \log(1+x^2) - 2x + 2 	an^{-1} x - \frac{1}{2} - \frac{x^2}{2}$.Comparing this with the given form $(x + \frac{x^2}{2} + \frac{1}{2}) \log(1+x^2) + g(x) + C$,we identify:$g(x) = -2x + 2 	an^{-1} x - \frac{x^2}{2}$.

If $\int(1+x) \log \left(1+x^2\right) d x=\left(x+\frac{x^2}{2}+\frac{1}{2}\right) \log \left(1+x^2\right)+g(x)+C$,then $g(x)=$

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