How can I solve $\int\limits_0^1\frac{\sqrt{x}}{(x+3)\sqrt{x+3}}dx$ without trigonometric substitution? I have the following integral to solve:
$$\int_0^1 \dfrac{\sqrt{x}}{(x+3)\sqrt{x+3}}dx$$
without using trigonometric substitution. My textbook gives me the following hint:
$$t = \sqrt{\dfrac{x}{x+3}}$$
But I don't see how this would help me. If I differentiate that, I get:
$$dt = \dfrac{1}{2} \cdot \sqrt{\dfrac{x+3}{x}} \cdot \dfrac{3}{(x+3)^2} dx$$
$$dt = \dfrac{3}{2(x+3)^2} \cdot \dfrac{1}{t} dx$$
And I'm stuck. If I substitute in the original integral, I'll have terms with both $x$ and $t$. So how can I use the given hint?

Solve for $x$:
$$t^2 = \frac{x}{x+3} = 1 - \frac{3}{x+3} \implies x = \frac{3}{1-t^2} - 3$$
then we have
$$dx = \frac{6t}{(1-t^2)^2}dt$$
and plugging in to the integral gets us
$$ \int_0^{\frac{1}{2}} \left(\frac{1-t^2}{3}\right)\cdot (t) \cdot \left(\frac{6t}{(1-t^2)^2}\right)dt = 2\int_0^{\frac{1}{2}} \frac{t^2}{1-t^2}dt = \int_0^{\frac{1}{2}} \frac{2}{1-t^2}-2dt$$
$$ = \int_0^{\frac{1}{2}} \frac{1}{1+t}+\frac{1}{1-t}-2dt = \log\left(\frac{1+t}{1-t}\right)-2t\Biggr|_0^{\frac{1}{2}} = \log(3)-1$$
