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?

$$\int_{3}^{4}\frac{\sqrt{x-3}}{x\sqrt{x}}\,dx\stackrel{x\mapsto 3x}{=}\int_{1}^{4/3}\frac{\sqrt{x-1}}{x\sqrt{x}}\,dx=\int_{1}^{4/3}\frac{1}{x}\sqrt{1-\frac{1}{x}}\,dx\stackrel{x\mapsto 1/x}{=}\int_{3/4}^{1}\frac{1}{x}\sqrt{1-x}\,dx $$
equals
$$ \int_{0}^{1/4}\frac{\sqrt{x}}{1-x}\,dx \stackrel{x\mapsto x^2}{=}\int_{0}^{1/2}\frac{2x^2}{1-x^2}\,dx=\int_{0}^{1/2}\left[-2+\frac{1}{1+x}+\frac{1}{1-x}\right]\,dx $$
which is easily seen to be $\log 3-1$.
