Integral of exponential function

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$\int \left(\sec13x\;\tan13x\right)\;e^{\sec13x} \; dx$

**Answer**

$\int \left(\sec13x\;\tan13x\right)\;e^{\sec13x} \; dx$

Substitute $u= \sec13x$. So, $du =13\sec13x\;\tan13x\; dx$

And, $\frac{1}{13}\; du = \sec 13x\;\tan13x\; dx$

Then $\int \left(\sec13x\;\tan13x\right)\;e^{\sec13x} \; dx$

$=\frac{1}{13}\int e^{u} \; du$

$=\frac{1}{13}e^u + C $

Substituting back $u= \sec 13x$

$=\frac{1}{13}e^{\sec 13x} + C$