Integrals look easy but I'm still confused

[daivinhtran]

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

Homework Statement

evaluate the integral ∫(36/(2x+1)^3)dx

Homework Equations

dx^n/dx = nx^(n-1)

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??

 

[daivinhtran]

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

 

[daivinhtran]

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

 

[pierce15]

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

 

[Mark44]

Homework Statement

evaluate the integral ∫(36/(2x+1)^3)dx

Homework Equations

dx^n/dx = nx^(n-1)

The Attempt at a Solution

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

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
$\int x^{-1}dx = ln|x| + C$

∫(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??