- #1
Mr Davis 97
- 1,462
- 44
Homework Statement
##\displaystyle \int \frac{x}{\sqrt{x + 1}}dx##
Homework Equations
The Attempt at a Solution
[/B]
First, I let ##u = x + 1##. Then ##du = dx## and ##x = u - 1##.
So ##\displaystyle \int \frac{u - 1}{\sqrt{u}}du = \int u^{\frac{1}{2}} - u^{-\frac{1}{2}}du = \frac{2}{3}u^{\frac{3}{4}} - 2u^{\frac{1}{2}} + C = \frac{2}{3}u\sqrt{u} - 2u + C = \frac{2}{3}u(\sqrt{u} - 3) + C = \frac{2}{3}(x + 1)(\sqrt{x + 1} - 3) + C##
However, the answer book says that the correct answer is ##\displaystyle \frac{2}{3}(x - 1)\sqrt{x + 1} + C##
What am I doing wrong?