How do you integrate int_0^1 1/(1+(x)^(1/2)) dx ?

Feb 14, 2018

Write ${x}^{\frac{1}{2}}$ as $\sqrt{x}$:

${\int}_{0}^{1} \frac{1}{1 + \sqrt{x}} \mathrm{dx}$

To do the indefinite integral, I used a tip from WolframAlpha :

Let $u = \sqrt{x}$ then $\mathrm{du} = \frac{1}{2 \sqrt{x}} \mathrm{dx}$

We need to solve for $\mathrm{dx}$:

$\mathrm{dx} = 2 \sqrt{x} \mathrm{du}$

But $\sqrt{x} = u$:

$\mathrm{dx} = 2 u \mathrm{du}$

Perform the substitutions:

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

Add 0 to the numerator in the form if $+ 1 - 1$

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

Separate into two fractions:

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

The first fraction becomes 1:

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

Both terms are trival to integrate:

$\int \frac{1}{1 + {\left(x\right)}^{\frac{1}{2}}} \mathrm{dx} = 2 \left(u - \ln \left(u + 1\right)\right) + C$

Reverse the substitution:

$\int \frac{1}{1 + {\left(x\right)}^{\frac{1}{2}}} \mathrm{dx} = 2 \left(\sqrt{x} - \ln \left(\sqrt{x} + 1\right)\right) + C$

To integrate from 0 to 1 we evaluate the right side at $x = 1$ and subtract the right side evaluated at $x = 0$:

${\int}_{0}^{1} \frac{1}{1 + {\left(x\right)}^{\frac{1}{2}}} \mathrm{dx} = \left\{2 \left(\sqrt{1} - \ln \left(\sqrt{1} + 1\right)\right) + C\right\} - \left\{2 \left(\sqrt{0} - \ln \left(\sqrt{0} + 1\right)\right) + C\right\}$

The constants of integration sum to 0 and $\sqrt{0} - \ln \left(1\right) = 0$, therefore, the answer is:

${\int}_{0}^{1} \frac{1}{1 + {\left(x\right)}^{\frac{1}{2}}} \mathrm{dx} = 2 - 2 \ln \left(2\right)$

Feb 14, 2018

$f \left(x\right) = \frac{1}{1 + \sqrt{x}}$

let $\sqrt{x} = t$
$x = {t}^{2}$
differentiating both the sides,
$\mathrm{dx} = 2 t . \mathrm{dt}$

$\int f \left(x\right) = \int \frac{1}{1 + \sqrt{x}} \mathrm{dx}$
$= \int \frac{1}{1 + t} 2 t . \mathrm{dt} = \int \frac{2 t}{1 + t} \mathrm{dt}$
applying simple polynomial division,
$\int \left(2 - \frac{2}{1 + t}\right) \mathrm{dt}$
$\int 2. \mathrm{dt} - \int \frac{2}{1 + t} \mathrm{dt}$
$2 \int \mathrm{dt} - 2 \int \frac{1}{1 + t} \mathrm{dt}$
$2 t - 2 \ln \left(1 + t\right) + c$

replacing, $t = \sqrt{x}$
$2 \sqrt{x} - 2 \ln \left(1 + \sqrt{x}\right) + c$

now, applying the limits,
$2 \sqrt{1} - 2 \ln \left(1 + \sqrt{1}\right) - \left(2 \sqrt{0} - 2 \ln \left(1 + \sqrt{0}\right)\right)$
$2 - 2 \ln \left(1 + 1\right) - \left(- 2 \ln \left(1\right)\right)$ as color(purple)(sqrt1=1; sqrt0=0;ln1=0
$2 - 2 \ln 2$