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
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