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Question

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

Vedclass · Accepted Answer

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

Vedclass · Answer

(B) To evaluate the integral $\int 	an^{-1} x \, dx$,we use the method of integration by parts: $\int u \, dv = uv - \int v \, du$. Let $u = 	an^{-1} x$ and $dv = dx$. Then $du = \frac{1}{1 + x^2} \, dx$ and $v = x$. Applying the formula: $\int 	an^{-1} x \, dx = x 	an^{-1} x - \int x \cdot \frac{1}{1 + x^2} \, dx$ $= x 	an^{-1} x - \int \frac{x}{1 + x^2} \, dx$ To solve $\int \frac{x}{1 + x^2} \, dx$,let $t = 1 + x^2$,so $dt = 2x \, dx$ or $x \, dx = \frac{1}{2} dt$. $= x 	an^{-1} x - \frac{1}{2} \int \frac{1}{t} \, dt$ $= x 	an^{-1} x - \frac{1}{2} \log|t| + C$ $= x 	an^{-1} x - \frac{1}{2} \log(1 + x^2) + C$.

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

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