# Thread: Is This Correct? Looks like a simple Intergral Problem...

1. ## Is This Correct? Looks like a simple Intergral Problem...

Evaluate:

Intergral of X Times squareroot 5X+1 DX
(see attached pic of problem)

So I used substitution,

U= 5X+1
Du= 5Dx
Dx= 1/5Du

Next I found Intergral UDU

U^1/2 Times 1/5DU =

2/3U^3/2 Times 1/5DU

2/15U^3/2 + C

Simplified to:

2/15(5x+1)^3/2 + C

Is that correct?

2. EDIT: wait... no, forgot the chain rule.

EDIT EDIT:
Take u = 5x + 1, so that x = (u - 1)/5 and du = 5 dx

=> the integral of (1/5) * (1/5) [(u-1)(sqrt(u)] du (substituting **x**, u and du)
=> the integral of (1/25) (u^3/2 - u^1/2) du
(1/25) * (2/5)u^5/2 - (2/3)u^3/2

3. Take u = 5x + 1, so that x = (u - 1)/5 and du = 5 dx

=> the integral of (1/5) * (1/5) [(u-1)(sqrt(u)] du (substituting **x**, u and du)
=> the integral of (1/25) (u^3/2 - u^1/2) du
(1/25) * (2/5)u^5/2 - (2/3)u^3/2

...yes this is correct

4. Hello,

Make $\displaystyle t=\sqrt{5x+1}$
$\displaystyle dt=\frac 52 \frac{dx}{\sqrt{5x+1}} \Rightarrow dx=\frac 25 ~ t ~dt$
And $\displaystyle t^2=5x+1 \Rightarrow x=\frac{t^2-1}{5}$

So the integral is now :
$\displaystyle \frac 25 \int t \left(\frac{t^2-1}{5}\right) ~dt=\frac{2}{25} \int t^3-t ~dt$