# Integrals look easy but I'm still confused

1. Nov 28, 2012

### daivinhtran

Integrals...look easy but I'm still confused :(

1. The problem statement, all variables and given/known data
evaluate the integral ∫(36/(2x+1)^3)dx

2. Relevant equations
dx^n/dx = nx^(n-1)

3. The attempt at a solution
∫(36/(2x+1)^3)dx = 6ln[(2x+1)^3]/((2x + 1)^2) ( I know this is wrong, but why??)

∫(36/(2x+1)^3)dx = -9/[(2x+1)^2)

I know the second one I did is right...but Why was the first one wrong??

2. Nov 28, 2012

### daivinhtran

Re: Integrals...look easy but I'm still confused :(

when I tried to differentiate 6ln[(2x+1)^3]/((2x + 1)^2) ...I get back to the same one though

3. Nov 28, 2012

### daivinhtran

Re: Integrals...look easy but I'm still confused :(

oh sorry, nevermind...I already found what I did wrong....

4. Nov 28, 2012

### pierce15

Re: Integrals...look easy but I'm still confused :(

You don't even need that natural log here, you just need a u substitution. If you were given this integral, I assume that you know what a u substitution is, but if you don't, tell me and I'll show you what's going on here.
$$\int \frac{36}{{2x+1)^3}}dx$$
$$36\int \frac{dx}{(2x+1)^3}$$
$$u=2x+1, du=2 dx \to dx=du/2$$
$$36\int \frac{du}{2u^3}$$
$$18\int u^{-3}du$$
$$18u^{-2}/-2+C$$
$$-9u^{-2}+C$$
$$-\frac{9}{(2x+1)^2}+C$$

5. Nov 28, 2012

### Staff: Mentor

Re: Integrals...look easy but I'm still confused :(

This is wrong because it is NOT true that
$$\int \frac{1}{f(x)}dx = ln|f(x)| + C$$

The correct formula is
$$\int \frac{1}{x}dx = ln|x| + C$$

Another way to write this is
$\int x^{-1}dx = ln|x| + C$