# Integral form

1. Apr 23, 2005

i don't see why the following integral:$$\int\frac{dx}{\sqrt{x}(x + 1)}$$ uses the arctan formula. i know how to solve integrals. i just don't see why or how you can rewrite it using the arctan formula. Isn't the arctan formula used for integrals of the form: $$\int\frac{du}{a^2 + u^2}=\frac{1}{a}arctan\frac{u}{a}$$

2. Apr 23, 2005

### p53ud0 dr34m5

$$\int \frac{dx}{\sqrt{x}(x+1)}$$
you have to use u-substitution.
so, you have:
$$u = \sqrt{x}~~~du=\frac{1}{2 \sqrt{x}$$
now, here's where it gets tricky:
$$x=u^2$$
now, you can substitute back in.
$$\int \frac{du}{u^2+1}$$
now, just integrate and plug u back in, and you are done.

*edit* it isn't let me post du...well, i cant see it from my comp, but du = 1/2sqrt(x)dx

Last edited: Apr 23, 2005
3. Apr 23, 2005

thanks i see now.

4. Apr 23, 2005

5. Apr 24, 2005

### dextercioby

He missed the 2 when differentiating $x=u^{2}$.

Daniel.

6. Apr 24, 2005

### Jameson

$$u = \sqrt{x}$$
$$du = \frac{1}{2}x^{-\frac{1}{2}}dx$$

??? What is wrong with that?

7. Apr 24, 2005

### dextercioby

Nothing,just that u didn't take it into account in the post i was referring to.

Daniel.

8. Apr 24, 2005

### p53ud0 dr34m5

hehe, yea i left out the 2...so its half the integral when you work it out. this computer at work was frustrating me, so i forgot it. (my excuse :rofl: )