Mathematics

# Solve:-$\displaystyle\int {\dfrac{{{e^x}(1 + x)}}{{{{\cos }^2}(x{e^x})}}} dx$

$I=\tan \left( x{{e}^{x}} \right)+C$

##### SOLUTION

Consider the given integral.

$I=\displaystyle\int{\dfrac{{{e}^{x}}\left( 1+x \right)}{{{\cos }^{2}}\left( x{{e}^{x}} \right)}}dx$

Let $t=x{{e}^{x}}$

$\dfrac{dt}{dx}=x{{e}^{x}}+{{e}^{x}}$

$dt={{e}^{x}}\left( x+1 \right)dx$

Therefore,

$I=\displaystyle\int{\dfrac{dt}{{{\cos }^{2}}t}}$

$I=\displaystyle\int{{{\sec }^{2}}tdt}$

$I=\tan t+C$

On putting the value of $t$, we get

$I=\tan \left( x{{e}^{x}} \right)+C$

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Single Correct Medium Published on 17th 09, 2020
Questions 203525
Subjects 9
Chapters 126
Enrolled Students 86

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