Integral of exponential function

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$\int \left({\sec}^2x\;\tanx\right)\;e^{{\tan}^2x} \; dx$

**Answer**

$\int \left({\sec}^2x\;\tanx\right)\;e^{{\tan}^2x} \; dx$

Substitute $u= {\tan}^2x$. So, $du =2{\sec}^2x\tanx\; dx$

And, $\frac{1}{2}\; du = {\sec}^2 x\;\tanx\; dx$

Then $\int \left({\sec}^2x\;\tanx\right)\;e^{{\tan}^2x} \; dx$

$=\frac{1}{2}\int e^{u} \; du$

$=\frac{1}{2}e^u + C $

Substituting back $u= {\tan}^2 x$

$=\frac{1}{2}e^{{\tan}^2x} + C$