I'm searching the antiderivative of rational functions :
1)$\int \frac {1+x}{\sqrt{2x+1}} dx$
For this one we have $t=\sqrt{2x+1}$ then $dt=\frac {1}{\sqrt {2x+2}} dx$ but then I do not see the way to compute the antiderivative.
Same thing for 2)$\int \frac {1}{x+\sqrt{x+1}}$
What are the methods to find antiderivative of rational functions like these ones ?
Thank you
For $(1)$: Try the substitution $u= 2x+1$. Then $du = 2\,dx\implies \frac 12\,du = dx$, and $x =\frac12(u-1) \implies x+1 = \frac12(u-1) + 1 = \frac 12 (u + 1)$ Then your integral becomes $$\int \frac {1+x}{\sqrt{2x+1}} dx = \frac 14\int \frac{u+1}{u^{1/2}}\,du\ =\frac 14\int \left(u^{1/2} + u^{-1/2}\right)\,du$$
Can your take it from here?
Using your substitution, $t=\sqrt{2x+1}$, so $dt = \frac{2dx}{2\sqrt{2x+1}}=\frac{dx}{\sqrt{2x+1}}$. On the other hand, $x=\frac{1}{2}(t^2-1)$, so the integral becomes $$ \int (1+x)\frac{dx}{\sqrt{2x+1}} = \int \left(1+\frac{1}{2}(t^2-1) \right) \, dt = \frac{1}{2}\int \left(1+t^2 \right) \, dt, $$ which I'm sure you can do.
For the second, again substitute for the square root: $y=\sqrt{x+1}$, so $x=y^2-1$, and $dx=2y \, dy$. You should find the integral reduces to $$ \int \frac{2y}{y^2+y-1} \, dy, $$ which you can then do in the usual way with partial fractions. (Not very nice partial fractions, unfortunately, but it does work.)
