Evaluate the integral: int operatorname{Tan}^ | vedclass

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Question

(D) Let $I = \int \operatorname{Tan}^{-1}\left(x^{\frac{1}{3}}\right) d x$. Substitute $x = t^3$,then $dx = 3t^2 dt$. $I = \int \operatorname{Tan}^{-1

Vedclass · Accepted Answer

$x \operatorname{Tan}^{-1}\left(x^{\frac{1}{3}}\right)+\frac{1}{2} \log \left(1+x^{\frac{2}{3}}\right)-\frac{1}{2} x^{\frac{2}{3}}+c$

Vedclass · Answer

(D) Let $I = \int \operatorname{Tan}^{-1}\left(x^{\frac{1}{3}}\right) d x$. Substitute $x = t^3$,then $dx = 3t^2 dt$. $I = \int \operatorname{Tan}^{-1}(t) \cdot 3t^2 dt = 3 \int t^2 \operatorname{Tan}^{-1}(t) dt$. Using integration by parts,let $u = \operatorname{Tan}^{-1}(t)$ and $dv = t^2 dt$. Then $du = \frac{1}{1+t^2} dt$ and $v = \frac{t^3}{3}$. $I = 3 \left[ \frac{t^3}{3} \operatorname{Tan}^{-1}(t) - \int \frac{t^3}{3(1+t^2)} dt \right] = t^3 \operatorname{Tan}^{-1}(t) - \int \frac{t^3}{1+t^2} dt$. Since $\frac{t^3}{1+t^2} = \frac{t(t^2+1)-t}{1+t^2} = t - \frac{t}{1+t^2}$,$I = t^3 \operatorname{Tan}^{-1}(t) - \int \left( t - \frac{t}{1+t^2} \right) dt = t^3 \operatorname{Tan}^{-1}(t) - \frac{t^2}{2} + \frac{1}{2} \log(1+t^2) + c$. Substituting $t = x^{\frac{1}{3}}$,we get $I = x \operatorname{Tan}^{-1}(x^{\frac{1}{3}}) - \frac{1}{2} x^{\frac{2}{3}} + \frac{1}{2} \log(1+x^{\frac{2}{3}}) + c$.

Evaluate the integral: $\int \operatorname{Tan}^{-1}\left(x^{\frac{1}{3}}\right) d x$

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