The correct option is A x tan^ 1 x 12 log | 1 + x^2| + c I=∫Itan^ 1x×II1dx From integration by parts, I=tan^ 1x· x ∫11+x^2x dx I_1=∫xx^2+1dx ...

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Continuity of a Function

∫ tan-1 xdx =

Question

$\int t a n^{-} 1 x d x =$

x t a n^{-} 1 x - \frac{1}{2} l o g | 1 + x^{2} | + c

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x t a n^{-} 1 x + \frac{1}{2} l o g | 1 + x^{2} | + c

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x t a n^{-} 1 x - \frac{1}{2} l o g | 1 + x | + c

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x t a n^{-} 1 x - \frac{1}{2} l o g | 1 + x^{3} | + c

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Solution

The correct option is A $x t a n^{-} 1 x - \frac{1}{2} l o g | 1 + x^{2} | + c$
$I = \int {tan}^{- 1} x I \times 1 I I d x$
From integration by parts,
$I = {tan}^{- 1} x \cdot x - \int \frac{1}{1 + x^{2}} x d x$
$I_{1} = \int \frac{x}{x^{2} + 1} d x$ Let $x^{2} + 1 = t$
$\Rightarrow 2 x d x = d t$
$= \int \frac{d t}{2 t} \Rightarrow I_{1} = \frac{1}{2} l n t + c = \frac{1}{2} l n | x^{2} + 1 | + c$
$\Rightarrow I = x {tan}^{- 1} x - \frac{1}{2} l n | x^{2} + 1 | + c$
$\Rightarrow (A)$ .

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