Integral of x*sec^2(x) dx

1. Use integration by parts:

   $$\int \operatorname{u} \operatorname{dv} = \operatorname{u}\operatorname{v} - \int \operatorname{v} \operatorname{du}$$

   Let $u{\left(x \right)} = x$ and let $\operatorname{dv}{\left(x \right)} = \sec^{2}{\left(x \right)}$.

   Then $\operatorname{du}{\left(x \right)} = 1$.

   To find $v{\left(x \right)}$:

   1. $$\int \sec^{2}{\left(x \right)}\, dx = \tan{\left(x \right)}$$

   Now evaluate the sub-integral.

2. Rewrite the integrand:

   $$\tan{\left(x \right)} = \frac{\sin{\left(x \right)}}{\cos{\left(x \right)}}$$

3. Let $u = \cos{\left(x \right)}$.

   Then let $du = - \sin{\left(x \right)} dx$ and substitute $- du$:

   $$\int \frac{1}{u}\, du$$

   1. The integral of a constant times a function is the constant times the integral of the function:

      $$\int \left(- \frac{1}{u}\right)\, du = - \int \frac{1}{u}\, du$$

      1. The integral of $\frac{1}{u}$ is $\log{\left(u \right)}$.

      So, the result is: $- \log{\left(u \right)}$

   Now substitute $u$ back in:

   $$- \log{\left(\cos{\left(x \right)} \right)}$$

4. Add the constant of integration:

   $$x \tan{\left(x \right)} + \log{\left(\cos{\left(x \right)} \right)}+ \mathrm{constant}$$

The answer is: