Consider $\int x {tan}^{- 1} x d x = A (x^{2} + 1) {tan}^{- 1} x + B x + C,$ where C is the constant of integration.
What is the value of A?

1

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\frac{1}{2}

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\frac{- 1}{2}

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\frac{1}{4}

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Solution

The correct option is B $\frac{1}{2}$
$\int x {tan}^{- 1} x d x$ = $\frac{x^{2}}{2} {tan}^{- 1} x - \frac{1}{2} \int \frac{x^{2}}{1 + x^{2}} d x$

$= \frac{x^{2}}{2} {tan}^{- 1} x - \frac{1}{2} x + {tan}^{- 1} x + C$

$= \frac{1}{2} (x^{2} + 1) tan - 1 x - \frac{1}{2} x + C$

hence $A = \frac{1}{2}$

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