Integrate $\frac{x-1}{\sqrt{x}+1}$ by power-law

Question

How to evaluate the following integral by power-law?

$$\int\dfrac{x-1}{\sqrt{x}+1}\,\mathrm dx$$

Here is my solution which is apparently wrong: $$\begin{align} \int\frac{x-1}{\sqrt x+1}\,\mathrm dx &= \int (x-1)(x^{1/2}+1)^{-1}\,\mathrm dx \\ &= \frac23 x\sqrt x+\frac12x^2 - 2x^{1/2}-x+c \end{align}$$

There should be a fairly easy solution to integrate this.

Answer 1

Note that $$\frac{x-1}{\sqrt{x}+1}=\sqrt{x}-1.$$

This will be clearer if you recall that $\frac{y^2-1}{y+1}=\frac{(y-1)(y+1)}{y+1}=y-1$. Put $y=\sqrt{x}$.

Remark: Your calculation seems to have used the incorrect "rule" $(a+b)^{-1}=a^{-1}+b^{-1}$.

Answer 2

HINT:

$$\int \frac{x-1}{\sqrt x+1}dx=\int (\sqrt x-1)dx$$

Now, use $\int x^m\ dx=\frac{x^{m+1}}{m+1}+C$ for real $m\ne-1$