Mathematics

# Evaluate the following integral$\displaystyle\int{\dfrac{1+\tan{x}}{x+\log{\sec{x}}}dx}$

##### SOLUTION
Let
$t=x+\log{\sec{x}}\Rightarrow\,dt=\left(1+\dfrac{1}{\sec{x}}\sec{x}\tan{x}\right)dx=\left(1+\tan{x}\right)dx$

$\displaystyle\int{\dfrac{1+\tan{x}}{x+\log{\sec{x}}}dx}$

$=\displaystyle\int{\dfrac{dt}{t}}$

$=\log{\left|t\right|}+c$

$=\log{\left|x+\log{\sec{x}}\right|}+c$ where $t=x+\log{\sec{x}}$

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Subjective Medium Published on 17th 09, 2020
Questions 203525
Subjects 9
Chapters 126
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