Integral of Secant function

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$\int \sec\left(4x-1\right)\; dx$

**Answer**

$\int \sec\left(4x-1\right)\; dx$

Substitute $u=4x-1$. So, $du =4\; dx$

Or, $\frac{1}{4}\; du = dx$

And, $\int \sec\left(4x-1\right)\; dx = \frac{1}{4}\int \sec u \;du$

$=\frac{1}{4}\ln \left|\sec u + \tan u \right| + C$

Substituting back $u=4x-1$

$=\frac{1}{4}\ln \left| \sec\left(4x-1\right)+ \tan\left(4x-1\right)\right|+C$