Evaluate the following integrals:


Evaluate: $\int \frac{\sec ^{2} \sqrt{x}}{\sqrt{x}} \mathrm{dx}$


let $\sqrt{x}=t$

Differentiating on both sides we get,

$\frac{1}{2 \sqrt{x}} d x=d t$

$\frac{1}{\sqrt{x}} d x=2 d t$

substituting it in $\int \frac{\sec ^{2} \sqrt{x}}{\sqrt{x}} d x$ we get,

$=\int 2 \sec ^{2} t d t$

$=2 \tan t+c$

$=2 \tan \sqrt{x}+c$

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