integral of $\frac{(1+x^{1/2})^9}{ x^{1/2}}$ I have taken the following steps to arrive at the conclusion of 4x(1+x^(1/2)^9 + C
$$u= 1+x^{1/2}$$
$$du= 1/2*x^{-1/2}dx$$
$$dx= 2x^{1/2}$$
$$u^9 / x^{1/2} dx$$
$$u^9*4x+C$$
$$4x(1+x^{1/2}))^9 +C$$
I am looking for confirmation if I did this problem correctly and some assistance if I missed a step somewhere. 
 A: Let $\sqrt x=t$ then $\frac12\frac{dx}{\sqrt x}=dt$ so
$$\int\frac{(1+\sqrt x)^9}{\sqrt x}dx=2\int(1+t)^9dt=\frac15(1+t)^{10}+C=\frac15(1+\sqrt x)^{10}+C$$
A: It's simpler than that:  the substitution you chose already has its derivative in the integrand:  if $u = 1+x^{1/2}$, then $du = \frac{1}{2}x^{-1/2} \, dx$, and it follows that $$\begin{align*} \int \frac{(1+x^{1/2})^9}{x^{1/2}} \, dx &= 2 \int (1+x^{1/2})^9 \cdot \frac{1}{2} x^{-1/2} \, dx = 2 \int u^9 \, du \\ &= 2 \cdot \frac{1}{10}u^{10} + C = \frac{(1+x^{1/2})^{10}}{5} + C. \end{align*}$$
A: One way to quickly confirm/disprove your solution is to look at the largest power of $x$.
If you were to multiply out the numerator of the integrand (possible, and not too tedious in this case) and divide by the denominator, the largest power of $x$ in the rearranged integrand would be $x^4$.  After integrating, the max power would be $x^5$.  In your proposed solution, the max power of $x$ is $x^{5.5}$...
Well, the other responses show the error...
It's interesting that Wolfram Alpha solves this integral by brute force;  it does expand the integrand and then integrates term by term.  Even if you change the exponent from $9$ to $109$, Wolfram just grinds it out.
But if you change the exponent to, say $12.5$, Wolfram is forced to get crafty and make the substitution suggested above... 
