int_0^1 tan^{-1} x , dx = | vedclass

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(B) To evaluate the integral $I = \int_0^1 	an^{-1} x \, dx$,we use integration by parts: $\int u \, dv = uv - \int v \, du$.Let $u = 	an^{-1} x$ an

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$\frac{\pi}{4} - \log \sqrt{2}$

Vedclass · Answer

(B) To evaluate the integral $I = \int_0^1 	an^{-1} x \, dx$,we use 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:$I = [x 	an^{-1} x]_0^1 - \int_0^1 \frac{x}{1+x^2} \, dx$.Evaluating the first part: $(1 \cdot 	an^{-1} 1) - (0 \cdot 	an^{-1} 0) = \frac{\pi}{4} - 0 = \frac{\pi}{4}$.For the second part,let $t = 1+x^2$,so $dt = 2x \, dx$ or $x \, dx = \frac{1}{2} dt$.When $x=0, t=1$; when $x=1, t=2$.$\int_0^1 \frac{x}{1+x^2} \, dx = \frac{1}{2} \int_1^2 \frac{1}{t} \, dt = \frac{1}{2} [\log |t|]_1^2 = \frac{1}{2} (\log 2 - \log 1) = \frac{1}{2} \log 2 = \log 2^{1/2} = \log \sqrt{2}$.Thus,$I = \frac{\pi}{4} - \log \sqrt{2}$.

$\int_0^1 \tan^{-1} x \, dx =$

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