Evaluate the algebraic indefinite integral $\int \frac{\sqrt{x+1}-\sqrt{x-1}}{\sqrt{x+1}+\sqrt{x-1}}dx$

Question

I have no idea how to solve it, can you show me how to start with this integral?

$$\int \frac{\sqrt{x+1}-\sqrt{x-1}}{\sqrt{x+1}+\sqrt{x-1}}dx$$

Thank you.

Answer 1

$$\int \frac{\sqrt{x+1}-\sqrt{x-1}}{\sqrt{x+1}+\sqrt{x-1}}dx$$

First you'd better expand by: $\color{green}{−\sqrt{x+1}+\sqrt{x−1}}$ to "clean" the denominator

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

Now you can apply the linearity: $$\int \frac{-2\sqrt{x-1}\sqrt{x+1}-2x}{2}dx=-\int \sqrt{x-1}\sqrt{x+1}dx-\int \:xdx$$

FIRST: $$\int \sqrt{x-1}\sqrt{x+1}dx=\int \:\sqrt{x^2-1}dx$$ Now substitute $\color{green}{x=\sec(y), y =arcsec(x),\frac{dx}{dy}=\sec\left(y\right)\tan\left(y\right)}$

$$\int \:\sqrt{x^2-1}dx=\int \:\sec\left(y\right)\sqrt{\sec^2\left(y\right)-1}\tan\left(y\right)dy$$

Now you can simplify using: $\color{green}{\sec ^2\left(y\right)-1=\tan ^2\left(y\right)}$ and rewrite as: $$\color{blue}{\int \sec ^3\left(y\right)-\sec \left(y\right)dy}$$ Now it's easy...

Amarildo $\phi$

Answer 2

Hint. You may first observe that, multiplying numerator and denominator by the conjugate expression, one has $$ \frac{\sqrt{x+1}-\sqrt{x-1}}{\sqrt{x+1}+\sqrt{x-1}}=\frac{(\sqrt{x+1}-\sqrt{x-1})^2}{2}=x-\sqrt{x^2-1} $$ then make the change of variable $$ x=\cosh u,\quad dx=\sinh u\: du, $$ getting $$ \int\frac{\sqrt{x+1}-\sqrt{x-1}}{\sqrt{x+1}+\sqrt{x-1}}\:dx=\frac{\cosh^2 u}2-\int \sinh^2 u\:du. $$

Answer 3

Hint If we define $u$ to be the entire integrand and solve for $x$ (which is often a passable idea if no more sensible option appears available), we find that $$x = \frac{u^2 + 1}{u},$$ and so $$\int \frac{\sqrt{x+1}-\sqrt{x-1}}{\sqrt{x+1}+\sqrt{x-1}} dx = \int u \cdot \frac{1}{2}\left( 1 - \frac{1}{u^2} \right) du . $$