Question

$integrate \: \: \: ({ \tan(y) }^{ - 1} \div 1 + {y}^{2} ) \: \times {e}^{ { \tan(y) }^{ - 1 } }$
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Answer 1

Solution:-

$\rm \implies \int \bigg( \dfrac{ \tan {}^{ - 1}y}{1 + {y}^{2} } \times {e}^{ \tan {}^{ - 1} y} \bigg) dy$

Using substitution method

$\rm let$

$\rm \implies \tan {}^{ - 1} y = t$

$\implies \rm \: \dfrac{d( \tan {}^{ - 1} y)}{dy} = \dfrac{dt}{dy}$

$\rm \implies \: \dfrac{1}{1 + {y}^{2} } = \dfrac{dt}{dy}$

$\rm \implies \: \dfrac{1}{1 + {y}^{2} } dy = dt$

Now put the value

$\rm \implies \int \bigg( \dfrac{ \tan {}^{ - 1}y}{1 + {y}^{2} } \times {e}^{ \tan {}^{ - 1} y} \bigg) dy$

$\rm \implies \int( t \times {e}^{t} \: )dt$

Now using Integration By part (UV method )

$\rm \implies \int(uv)dx = u \int vdx - \int \bigg\{ \bigg( \frac{du}{dx} \bigg) \int v \: dx \bigg\}dx$

SO let

$\rm u = t \: \: and \: \: {e}^{t} = v$

Now

$\rm \implies \: t \int {e}^{t} dt - \int \bigg( \dfrac{dt}{dt} \bigg( \int {e}^{t} dt \bigg) \bigg)dt$

$\implies \rm \: t {e}^{t} - \int(1 \times {e}^{t} )dt$

$\implies \rm \: t {e}^{t} - {e}^{t} + c$

Now put the value of t

$\rm \implies \: \tan {}^{ - 1} y \times {e}^{ \tan {}^{ - 1} y} - {e}^{ \tan {}^{ - 1} y} + c$

Answer

$\rm \implies \: \tan {}^{ - 1} y \times {e}^{ \tan {}^{ - 1} y} - {e}^{ \tan {}^{ - 1} y} + c$

$\rm \implies {e}^{ \tan {}^{ - 1} y } ( \tan {}^{ - 1}y - 1) + c$