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Q.

$\int x^{4} d (\cot^{- 1} x) d x = g (x) + C then g (1) = - - -$

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a

$\frac{π}{4} + \frac{2}{3}$

b

0

c

$\frac{2}{3} - \frac{π}{4}$

d

None

answer is B.

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Detailed Solution

$\begin{array}{l} I = \int x^{4} d (\cot^{- 1} x) d x \\ I = - \int \frac{x^{4}}{1 + x^{2}} d x \\ = - \int \frac{x^{4} - 1 + 1}{1 + x^{2}} d x \end{array}$

$\begin{array}{l} = - \int x^{2} - 1 + \frac{1}{1 + x^{2}} d x \\ = - \frac{x^{3}}{3} + x + \cot^{- 1} x \\ g (x) = \cot^{- 1} x - \frac{x^{3}}{3} + x \\ g (1) = \frac{π}{4} + \frac{2}{3} \end{array}$

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∫x4dcot−1⁡xdx=g(x)+C then g(1)=−−−