I=∫Tan−1⁡x1/31dx=Tan−1⁡x1/3∫1dx−∫11+x2/313x−2/3xdx=xTan−1⁡x1/3−13∫x1/31+x2/3dxPut 1+x2/3=t. Then 23x−1/3dx=dt⇒x−1/3dx=32dt∴∫x1/31+x2/3dx=∫x2/31+x2/3x−1/3dx=32∫t−1tdt=32(t−log⁡t)=321+x2/3−log⁡1+x2/3∴∫Tan−1⁡x1/3dx=xTan−1⁡x1/3−13321+x2/3−32log⁡1+x2/3+c=xTan−1⁡x1/3+12log⁡1+x2/3−12x2/3+c

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$\int {Tan}^{- 1} (x^{1 / 3}) d x =$

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$\begin{aligned} \frac{1}{2} \log (1 + x^{2 / 3}) - \frac{1}{2} x^{2 / 3} + c \end{aligned}$

$\begin{aligned} x {Tan}^{- 1} (x^{1 / 3}) - \frac{1}{2} x^{2 / 3} + c \end{aligned}$

$\frac{1}{2} \log (1 + x^{1 / 3}) - \frac{1}{2} x^{2 / 3} + c$

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

answer is D.

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Detailed Solution

$\begin{array}{l} I = \int {Tan}^{- 1} (x^{1 / 3}) 1 d x = {Tan}^{- 1} (x^{1 / 3}) \int 1 d x - \int \frac{1}{1 + x^{2 / 3}} \frac{1}{3} x^{- 2 / 3} x d x \\ = x {Tan}^{- 1} (x^{1 / 3}) - \frac{1}{3} \int \frac{x^{1 / 3}}{1 + x^{2 / 3}} d x \end{array}$

Put $1 + x^{2 / 3} = t . Then \frac{2}{3} x^{- 1 / 3} d x = d t \Rightarrow x^{- 1 / 3} d x = \frac{3}{2} d t$

$\begin{array}{l} ∴ \int \frac{x^{1 / 3}}{1 + x^{2 / 3}} d x = \int \frac{x^{2 / 3}}{1 + x^{2 / 3}} x^{- 1 / 3} d x = \frac{3}{2} \int \frac{t - 1}{t} d t = \frac{3}{2} (t - \log t) = \frac{3}{2} [(1 + x^{2 / 3}) - \log (1 + x^{2 / 3})] \\ ∴ \int {Tan}^{- 1} (x^{1 / 3}) d x = x {Tan}^{- 1} (x^{1 / 3}) - \frac{1}{3} [\frac{3}{2} (1 + x^{2 / 3}) - \frac{3}{2} \log (1 + x^{2 / 3})] + c \\ = x {Tan}^{- 1} x^{1 / 3} + \frac{1}{2} \log (1 + x^{2 / 3}) - \frac{1}{2} x^{2 / 3} + c \end{array}$