∫131(tan−1x)31+x2 dx=

∫π/6π/4t3.dt Tan−1x=t =(t44)π/6π/4 =14(π444−π464)

∫131(tan−1x)31+x2 dx=

∫π/6π/4t3.dt Tan−1x=t =(t44)π/6π/4 =14(π444−π464)

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$\int_{\frac{1}{\sqrt{3}}}^{1} \frac{{(\tan^{- 1} x)}^{3}}{1 + x^{2}} d x =$

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$\frac{1}{4} {(\frac{π}{12})}^{6}$

$\frac{1}{4} {(\frac{π}{12})}^{4}$

$\frac{1}{4} ({(\frac{π}{4})}^{4} - {(\frac{π}{6})}^{4})$

${(\frac{π}{8})}^{4}$

answer is C.

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Detailed Solution

$\int_{π / 6}^{π / 4} t^{3} . d t$ $T a n^{- 1} x = t$

$= {(\frac{t^{4}}{4})}_{π / 6}^{π / 4}$

$= \frac{1}{4} (\frac{π^{4}}{4^{4}} - \frac{π^{4}}{6^{4}})$

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∫131(tan−1x)31+x2 dx=

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$\int_{\frac{1}{\sqrt{3}}}^{1} \frac{{(\tan^{- 1} x)}^{3}}{1 + x^{2}} d x =$