int x tan^{-1} x , dx = | vedclass

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Question

(A) To solve the integral $\int x 	an^{-1} x \, dx$,we use the method of integration by parts: $\int u \, dv = uv - \int v \, du$.Let $u = 	an^{-1}

Vedclass · Accepted Answer

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

Vedclass · Answer

(A) To solve the integral $\int x 	an^{-1} x \, dx$,we use the method of integration by parts: $\int u \, dv = uv - \int v \, du$.Let $u = 	an^{-1} x$ and $dv = x \, dx$.Then $du = \frac{1}{1+x^2} \, dx$ and $v = \frac{x^2}{2}$.Applying the formula:$\int x 	an^{-1} x \, dx = \frac{x^2}{2} 	an^{-1} x - \int \frac{x^2}{2(1+x^2)} \, dx$$= \frac{x^2}{2} 	an^{-1} x - \frac{1}{2} \int \frac{x^2 + 1 - 1}{1+x^2} \, dx$$= \frac{x^2}{2} 	an^{-1} x - \frac{1}{2} \int (1 - \frac{1}{1+x^2}) \, dx$$= \frac{x^2}{2} 	an^{-1} x - \frac{1}{2} (x - 	an^{-1} x) + c$$= \frac{x^2}{2} 	an^{-1} x - \frac{1}{2}x + \frac{1}{2} 	an^{-1} x + c$$= \frac{1}{2}(x^2 + 1) 	an^{-1} x - \frac{1}{2}x + c$.

$\int x \tan^{-1} x \, dx = $

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