Integrating $\int{\left( x+2 \right)\sqrt{x-1}dx}$ I’m stuck on this relatively simple indefinite integral but, unfortunately, cannot figure out why my answer is wrong.
Here it is:
\begin{align}
  \int{\left( x+2 \right)\sqrt{x-1}dx}&=\int{\left( \left( x-1 \right)+3 \right)\frac{3}{2}\frac{1}{{{\left( \sqrt{x-1} \right)}^{2}}}\left( \frac{2}{3}{{\left( \sqrt{x-1} \right)}^{3}} \right)dx}\\&= 
 \int{\left( \left( x-1 \right)+3 \right)\frac{3}{2}\frac{1}{{{\left( \sqrt{x-1} \right)}^{2}}}{{\left( \sqrt{x-1} \right)}_{x}}'dx}\\&=\frac{3}{2}\int{\left( \left( x-1 \right)+3 \right)\frac{1}{x-1}d\sqrt{x-1}}\\&= 
 \frac{3}{2}\int{\left( 1+\frac{3}{x-1} \right)d\sqrt{x-1}}\\&=\frac{3}{2}\int{1d\sqrt{x-1}}+\frac{9}{2}\int{\frac{1}{{{\left( \sqrt{x-1} \right)}^{2}}}d\sqrt{x-1}}\\&=\frac{3}{2}\sqrt{x-1}-\frac{9}{2\sqrt{x-1}}
\end{align}
I'm fully aware there’re more compact methods to solve this but I’m interested to find the error in the solution given above.

If you differentiate $\sqrt{x-1}$, what you get is $\dfrac1{2\sqrt{x-1}}$ rather than $\displaystyle\frac23\sqrt{x-1}^3$. If you want to get $\displaystyle\frac23\sqrt{x-1}^3$, then you should differentiate $\dfrac4{15}\sqrt{x-1}^5$.
