I want to integrate this fraction using t-substitution:
$\int \frac{1+\sqrt{x+1}}{1-\sqrt{x+1}}dx$
$x\gt-1, $t=\sqrt{x+1}$
My attempt: $dx = \frac{dt}{\frac{1}{2\sqrt{x+1}} }$
$\int \frac{1+t}{1-t}dx$
$\int \frac{(1+t)\times 2 \sqrt{x+1 }}{1-t}dt$
$\int \frac{2 \sqrt{x+1 }+2t \sqrt{x+1 }}{1-t}dt$
$2 \int \frac{ \sqrt{x+1 }}{1-t}dt + 2 \int \frac{t\sqrt{x+1 }}{1-t} dt$
I'm not sure how to continue. I think the antiderivative of $\sqrt{x+1}$ is $\frac{2{(x+1)}^\frac 32}{3}$. But how do I handle the denominator?