Integration By Substitution

1. Aug 2, 2008

Air

1. The problem statement, all variables and given/known data:
using the substitution $u=3x+4$, work out:
$\int 2x \sqrt{3x+4}$

2. The attempt at a solution:
$\int 2x \sqrt{3x+4}$
$u=3x+4 \rightarrow \mathrm{d}x = \frac{\mathrm{d}u}{3}$
$\int \left( \frac{u-4}{3} \right)\left(u^{\frac12}\right) \ \frac{\mathrm{d}u}{3}$
$\frac19 \int u^{\frac32} - 4u^{\frac12} \ \mathrm{d}u$
$=\frac19 \left[ \frac{2u^{\frac52}}{5} - \frac{8u^{\frac32}}{3}\right] + c$
$=\frac{2u^{\frac52}}{45} - \frac{8u^{\frac32}}{27}}+c$
$=\frac{2(3x+4)^{\frac52}}{45} - \frac{8(3x+4)^{\frac32}}{27} + c$

3. The problem that I am encountering:
I was close to the answer but it is incorrect. The correct answer is:
$\frac{4(3x+4)^{\frac52}}{45} - \frac{16(3x+4)^{\frac32}}{27} + c$
Where have I gone wrong? Thanks in advance.

2. Aug 2, 2008

d_leet

You forgot to keep the 2 that was in front of the original problem after you made the substitution.

3. Aug 2, 2008

viciousp

Yeah looks like you did not distrbute the 2 when you set x=(u-4)/3. So change your 4th step to a 2/9 outside the integral.