# How do you find the integral of int 3/(2sqrtx(1+x)?

Nov 8, 2016

$3 \arctan \left(\sqrt{x}\right) + C$

#### Explanation:

$I = \int \frac{3}{2 \sqrt{x} \left(1 + x\right)} \mathrm{dx} = \frac{3}{2} \int \frac{\mathrm{dx}}{\sqrt{x} \left(1 + x\right)}$

If the substitution presented by Eric Sia is not immediately apparent, another substitution we can try is $u = \sqrt{x}$. This implies that $\mathrm{du} = \frac{1}{2 \sqrt{x}} \mathrm{dx}$ which we already have in the integrand.

$I = 3 \int \frac{1}{1 + x} \frac{1}{2 \sqrt{x}} \mathrm{dx} = 3 \int \frac{1}{1 + {u}^{2}} \mathrm{du}$

At this point, you may recognize that this is the arctangent integral. However, since this is in the "Integration by Trigonometric Substitution" section, we can derive the arctangent integral here using a trig substitution. Let $u = \tan \theta$. Thus $\mathrm{du} = {\sec}^{2} \theta d \theta$.

$I = 3 \int \frac{1}{1 + {\tan}^{2} \theta} \left({\sec}^{2} \theta d \theta\right) = 3 \int d \theta = 3 \theta + C$

Note that we used $1 + {\tan}^{2} \theta = {\sec}^{2} \theta$. Also, reverse the substitution of $u = \tan \theta$ to show that $\theta = \arctan \left(u\right)$. Since $u = \sqrt{x}$, we see that $\theta = \arctan \left(\sqrt{x}\right)$.

$I = 3 \arctan \left(\sqrt{x}\right) + C$