Integration by substitution - $\int (x+1)\sqrt{5x-6} \text{dx}$ $\int (x+1)\sqrt{5x-6} \text{dx}$
My working so far:
$u=5x-6$
$du=5dx$
$\int\dfrac{x+1}{5}\sqrt u \ \text{du}$
Substituting x
$\int \dfrac{(u+6)/5)+(5/5)}{5}\cdot\sqrt u \ \text{du}$
$\int (u+11)\sqrt u \ \text{du}$
I've got the same powers right when I integrate that, but I have my coefficients wrong. The answer is:
$\dfrac{2}{75}(5x-6)^{5/2}+\dfrac{22}{75}(5x-6)^{3/2}+c$
Where am I going wrong?
 A: Let the integral in question be
\begin{align}
I = \int (x+1) \sqrt{5x-6} \ dx
\end{align}
then making the substitution $u = 5x-6$ the integral becomes
\begin{align}
I &= \int \left( \frac{u+11}{5} \right) \ \sqrt{u} \ \frac{du}{5} \\
&= \frac{1}{25} \int \left[ u^{3/2} + 11 u^{1/2} \right] du \\
&= \frac{1}{25} \left[ \frac{2}{5} \ u^{5/2} + \frac{22}{3} u^{3/2} \right] 
\end{align}
and upon back substitution the result becomes
\begin{align}
I = \frac{2}{125} (5x-6)^{5/2} + \frac{22}{75} (5x-6)^{3/2} + c.
\end{align}
A: It looks like you made a simple cancellation error:  Your key simplification seems to equate
$${((u+6)/5)+(5/5)\over5}=u+11$$
(minor aside:  I added an extra parenthesis to "$(u+6)/5)$").  I'm guessing your thinking here was that the expression on the left hand side simplifies to 
$${(u+11)/5\over5}$$
and then that $5$ in the "numerator" cancels with the $5$ in the denominator.  The problem is, that top $5$ is actually in the denominator of the numerator, so it more properly just goes down and joins (multiplicatively) the other $5$, making the correct result
$${u+11\over25}$$
One other note:  You gave the "answer" with $2/75$ as one of its coefficients; the correct coefficient, as Leucippus found, is $2/125$.
