Question

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Answer 1

$\huge{\mathcal{\underline{\green{SOLUTION}}}} </p><p></p><p>$

$Let \: \: I \: = \displaystyle \int\limits \frac{ {tan}^{ - 1}x }{1 + \frac{1}{ {x}^{2} } } \, dx </p><p>$

$= \displaystyle \int\limits \frac{ {x}^{2} {tan}^{ - 1}x }{1 + {x}^{2} } \, dx$

$Let \: x \: = tan \theta$

$Then \: dx \: = {sec}^{2} \theta \: d \theta$

So

$I \: = \displaystyle \int\limits \frac{ {tan}^{2} \theta {tan}^{ - 1}(tan \theta) }{{sec}^{2} \theta } \, {sec}^{2} \theta \: d \theta$

$= \displaystyle \int\limits \theta \ {tan}^{2} \theta \: d \theta$

$= \displaystyle \int\limits \theta \ ({sec}^{2} \theta - 1) \: d \theta$

$= \displaystyle \int\limits \theta \ {sec}^{2} \theta \: d \theta - \displaystyle \int\limits \theta \ \: d \theta$

$= \theta \: tan \theta \: - log( |sec \theta| ) - \frac{ { \theta}^{2} }{2} + c$

$= x {tan}^{ - 1} x \: - log( | \sqrt{1 + {x}^{2} } | ) - \: \frac{ ( {{tan}^{ - 1} x)}^{2} }{2} + c$

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Answer 2

Letx=tanθ

$Then \: dx \: = {sec}^{2} \theta \: d \thetaThendx=sec2θdθ$

$</p><p>∫θ sec2θdθ−∫θ dθ</p><p>= \theta \: tan \theta \: - log( |sec \theta| ) - \frac{ { \theta}^{2} }{2} + c=θtanθ−log(∣secθ∣)−2θ2+c</p><p>$