Calculus find out if integral converges or diverges So I have this math problem, where I am supposed to find out whether or not the integral converges or diverges and solve.
$$\int_0^1 \frac{3\,dx}{\sqrt{x}(x+1)}$$
I'm not 100% sure as to figure out whether it converges or diverges with no calculator.
 A: It does, since your integrated function is equivalent to $\frac{1}{\sqrt x}$ when x->0, and this function is integrable around 0 (primitive = $2*\sqrt x$ -> 0 when x->0). On the rest of the interval the function is continuous so your function is integrable on [0,1]
As for the calculus, you can start with a change of variable:
u = $\sqrt x$  -> du = $\frac{1}{2*\sqrt x}*dx$
=> I = $2*\int_0^1 \frac{,du}{(u^2+1)}$  = 2*(arctan(1) - arctan(0)) = $\frac{\pi}{2}$
A: The only difficulty is at $x=0$.  At that point this function has a vertical asymptote.
$$
\frac{3}{x+1}\cdot\frac 1{\sqrt{x}}
$$
Since $3/(x+1)$ remains between $3$ and $3/2$ when $x$ is between $0$ and $1$, we have
$$
\frac 3 2 \cdot \frac 1 {\sqrt{x}} \le (\text{this function}) \le  3\cdot\frac 1 {\sqrt{x}}
$$
The integral either converges to a finite number or diverges to $\infty$.
The question of whether $\displaystyle\frac 3 2 \int_0^1 \frac{dx}{\sqrt{x}}$ converges and the question of whether $\displaystyle 3 \int_0^1 \frac{dx}{\sqrt{x}}$ converges are really both the same question, and the integral we're faced with is squeezed between them.
A: You can say $\displaystyle\int_{0}^{1}\frac{3}{\sqrt{x}(x+1)}dx=3\lim_{t\to0^{+}}\int_{t}^{1}\frac{1}{\sqrt{x}(x+1)}dx$.  
Now let $u=\sqrt{x}, x=u^2, dx=2udu$ to get
$6\displaystyle\lim_{t\to0^{+}}\int_{\sqrt{t}}^1\frac{1}{u^2+1}du=6\lim_{t\to0^{+}}\left[\arctan u\right]_{\sqrt{t}}^{1}=6\lim_{t\to0^{+}}\left(\frac{\pi}{4}-\arctan\sqrt{t}\right)=\frac{3\pi}{2}$.
