# How do you integrate?

## int_0^3dx/((x+2)(sqrtx+1)

Mar 26, 2018

$= - \frac{2}{3} \ln \left(1 + \sqrt{3}\right) + \frac{1}{3} \ln \left(\frac{5}{2}\right) + 2 \frac{\sqrt{2}}{3} {\tan}^{-} 1 \left(\sqrt{\frac{3}{2}}\right) \approx 0.47$

#### Explanation:

Substitute $x = {u}^{2}$. Then $\mathrm{dx} = 2 u \mathrm{du}$ and the integral becomes

${\int}_{0}^{3} \frac{\mathrm{dx}}{\left(x + 2\right) \left(\sqrt{x} + 1\right)} = {\int}_{0}^{\sqrt{3}} \frac{2 u \mathrm{du}}{\left({u}^{2} + 2\right) \left(u + 1\right)}$

We will now evaluate this integral using partial fractions. to thsi end, let us start with

$\frac{2 u}{\left({u}^{2} + 2\right) \left(u + 1\right)} = \frac{A}{u + 1} + \frac{B u + C}{{u}^{2} + 2} q \quad \implies$

$2 u = A \left({u}^{2} + 2\right) + \left(B u + C\right) \left(u + 1\right) = \left(A + B\right) {u}^{2} + \left(B + C\right) u + 2 A + C$

Comparing coefficients on both sides we get

$A + B = 0 , q \quad B + C = 2 , q \quad 2 A + C = 0$

This can be easily solved to get

$A = - \frac{2}{3} , \quad B = \frac{2}{3} , \quad \text{and } C = \frac{4}{3}$

Thus

$\frac{2 u}{\left({u}^{2} + 2\right) \left(u + 1\right)} = - \frac{2}{3} \frac{1}{u + 1} + \frac{1}{3} \frac{2 u + 4}{{u}^{2} + 2}$

and so

$\int \frac{2 u \mathrm{du}}{\left({u}^{2} + 2\right) \left(u + 1\right)} = - \frac{2}{3} \int \frac{\mathrm{du}}{u + 1} + \frac{1}{3} \int \frac{\left(2 u + 4\right) \mathrm{du}}{{u}^{2} + 2}$
$q \quad = - \frac{2}{3} \ln \left(u + 1\right) + \frac{1}{3} \ln \left({u}^{2} + 2\right) + \frac{4}{3 \sqrt{2}} {\tan}^{-} 1 \left(\frac{u}{\sqrt{2}}\right) + C$

Thus, the definite integral is

${\left(- \frac{2}{3} \ln \left(u + 1\right) + \frac{1}{3} \ln \left({u}^{2} + 2\right) + \frac{4}{3 \sqrt{2}} {\tan}^{-} 1 \left(\frac{u}{\sqrt{2}}\right)\right)}_{0}^{\sqrt{3}}$
$= - \frac{2}{3} \ln \left(1 + \sqrt{3}\right) + \frac{1}{3} \ln \left(\frac{5}{2}\right) + 2 \frac{\sqrt{2}}{3} {\tan}^{-} 1 \left(\sqrt{\frac{3}{2}}\right)$