# Thread: finding the integral

1. ## finding the integral

Find the integral of:

x * sqrt(5x^2 - 4) * dx

2. Let $u = 5x^2 - 4$

$du = 10x dx$

OR

$x dx = \frac{du}{10}$

3. would it be:

1/10 * (5x^2 - 4)^(3/2) + c

4. Originally Posted by DINOCALC09
would it be:

1/10 * (5x^2 - 4)^(3/2) + c
Almost, you forgot to divide by the three halves.

5. Originally Posted by DINOCALC09
Find the integral of:

x * sqrt(5x^2 - 4) * dx
Make-up

$\int {x\sqrt {5x^2 - 4} \,dx} = \frac{1}
{5}\int {5x\sqrt {5x^2 - 4} \,dx} .$

Set $u^2=5x^2-4\implies u\,du=5x\,dx,$

$\int {x\sqrt {5x^2 - 4} \,dx} = \frac{1}
{5}\int {u^2 \,du} = \frac{u^3 }
{{15}} + k = \frac{{\left( {5x^2 - 4} \right)^{3/2} }}
{{15}} + k.$