Calculus 2 integral $\int {\frac{2}{x\sqrt{x+1}}}\, dx$

Question

Make a substitution to express the integrand as a rational function and then evaluate the integral

$$\int {\frac{2}{x\sqrt{x+1}}}\, dx$$

What is the substitution that I have to make?

Answer 1

Let $u = \sqrt{x+1}$. Then $u^2 = x+1 \iff x = u^2 - 1$, and $dx = 2u\,du$. That gives you the integral

$$\int \dfrac{2u\,du}{u(u^2 - 1)} = 2\int \dfrac{du}{u^2 - 1}$$

Now we have our rational function. To evaluate, use the trigonometric substitution $u = \sec\theta$, and use the identity $$\tan^2 \theta + 1 = \sec^2\theta \iff \sec^2\theta - 1 = \tan^2\theta.$$

Answer 2

Your aim should be to get rid of the square root.

Warning: Spoiler below. :)

Try $u=\sqrt{x+1}$ so that $u^2-1=x$ Then $du = \frac{1}{2\sqrt{x+1}}\,dx = \frac{1}{2u}\,dx$ thus $2u\,du=dx$.