Integration of fraction with square root

Question

I have a problem with integrating of fraction $$ \int \frac{x}{x^2 + 7 + \sqrt{x^2 + 7}} $$ I have tried to rewrite it as $\int \frac{x^3 + 7x - x \sqrt{x^2 + 7}}{x^4 + 13x^2 + 42} = \int \frac{x^3 + 7x - x \sqrt{x^2 + 7}}{(x^2 + 6)(x^2 + 7)}$ and then find some partial fractions, but it wasn't succesful.

Answer 1

Since $x^2 + 7$ shows up twice it would make a good first substitution. If $u = x^2 + 7$ then $$\int \frac{x}{x^2 + 7 + \sqrt{x^2 + 7}} \, dx = \frac 12 \int \frac{1}{u + \sqrt{u}} \, du.$$ Can you take it from there?

Answer 2

Consider the integral \begin{align} I = \int \frac{x}{x^{2} + 7 + \sqrt{x^2 + 7}} \, dx \end{align} Make the substitution $u = \sqrt{x^{2} + 7}$ for which $x = \sqrt{u^{2} - 7}$ and the integral becomes \begin{align} I &= \int \frac{\sqrt{u^{2}-7}}{u^{2} + u} \cdot \frac{u \, du}{\sqrt{u^{2}-7}} \\ &= \int \frac{du}{1+u} \\ &= \ln(1+u) \end{align} Making the reverse substitution yields \begin{align} \int \frac{x}{x^{2} + 7 + \sqrt{x^2 + 7}} \, dx = \ln(1 + \sqrt{x^{2} + 7}). \end{align}