Integral help here please? I have to solve this 
$$\int \frac{dx}{(x+1)\sqrt{x^2+2x}}$$
First  make the substitution $u=x+1$ and get 
$$\int \frac{ds}{u\sqrt{u^2-1}}$$ 
Next I substitute $s=\sqrt{u^2-1}$  and I get $ds=u\,du/\sqrt{u^2-1}$. 
I have integral of $1/s^2+1\dots$ I replace for $s$ and I get the integral as
$$\arctan\sqrt{u^2-1} \, .$$
The answer on my textbook is $-\arcsin[1/(x+1)]$

You wanted $\;s=\sqrt{u^2-1}$, so $\;s^2=u^2-1$, $\;2s\,ds=2u\,du$, and $\;s\,ds=u\,du$.
Substituting gives us:
$$
\begin{align}\int\frac{du}{u\sqrt{u^2-1}} &= \int\frac{u\,du}{u^2 \sqrt{u^2-1}} \\&= \int\frac{s\,ds}{(s^2+1)s}\\&= \int\frac{ds}{s^2+1} \\ \\&= \arctan s + C.
\end{align}
$$
Now don't forget to "back substitute": $s = \sqrt{u^2-1} = \sqrt{(x+1)^2 - 1} = \sqrt{x^2 + 2x}$

So our answer is: $$\quad\arctan(\sqrt{x^2 + 2x}) + C$$

Note that with trig substitution, and trigonometric solutions, there can be any number of solutions (constants may vary), as is evident by the number of trigonometric identities! E.g.:
$$\arctan\sqrt{x^2+2x}-\left(\arcsin\left(\frac1{x+1}\right)-\frac\pi2\right) = 0$$
$$-\arcsin\frac1{x+1}=\arctan\sqrt{x^2+2x}-\text{ constant} = \arctan \sqrt{x^2 + 2x} + C$$
To check your integration answers, you can always take the the derivative of your solutions, and if you can derive the original integrand, you are "good to go." Try differentiating each solution (your's and the text's) and both will give you the equivalent of your original integrand.
