# How do you evaluate the integral int (1-sqrtx)/(1+sqrtx)?

Feb 23, 2017

$- x + 4 \left(1 + \sqrt{x}\right) - 4 \ln | 1 + \sqrt{x} | + C$

#### Explanation:

Use partial fractions to get rid of the radical in the numerator.

$\frac{A}{1} + \frac{B}{1 + \sqrt{x}} = \frac{1 - \sqrt{x}}{1 + \sqrt{x}}$

$A \left(1 + \sqrt{x}\right) + B = 1 - \sqrt{x}$

$A + A \sqrt{x} + B = 1 - \sqrt{x}$

This means that $\left\{\begin{matrix}A = - 1 \\ A + B = 1\end{matrix}\right.$

Solving, we get that $A = - 1$ and $B = 2$. The integral becomes

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

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

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

Let $u = 1 + \sqrt{x}$. Then $\mathrm{du} = \frac{1}{2 \sqrt{x}} \mathrm{dx}$ and $\mathrm{dx} = 2 \sqrt{x} \mathrm{du}$. We can also conclude that $2 \sqrt{x} = 2 \left(u - 1\right)$, because $\sqrt{x} = u - 1$ and we have two of those.

$\int - 1 \mathrm{dx} + 2 \int \frac{1}{u} \cdot 2 \left(u - 1\right) \mathrm{du}$

$\int - 1 \mathrm{dx} + 4 \int \frac{u - 1}{u} \mathrm{du}$

$\int - 1 \mathrm{dx} + 4 \int \frac{u}{u} \mathrm{du} - 4 \int \frac{1}{u} \mathrm{du}$

$\int - 1 \mathrm{dx} + 4 \int 1 \mathrm{du} - 4 \int \frac{1}{u} \mathrm{du}$

$- x + 4 u - 4 \ln | u | + C$

$- x + 4 \left(1 + \sqrt{x}\right) - 4 \ln | 1 + \sqrt{x} | + C$

Hopefully this helps!