# Thread: Solving Integral using substitution

1. ## Solving Integral using substitution [More help needed]

I have been having a trouble with this problem:

X / (4X^2+1)^2 dx<-------This is an integral that I have to solve using substitution

This is what I did.

u=4x^2+1
u'=8x
du/8=dx

So then I went back to original problem and plugged that in and got.

1/8*(1/(1/3)u^3) I got that from taking the anti derivative of x^2
I plugged u back in and got

1/8 * 1/(1/3)(4x^2+1)^3

after i simplify i get:
1/ ((8/3)(64x^6+1))

I would appreciate it if someone could tell me where i messed up. Thanks!

2. Hi !

Originally Posted by PensFan10
I would appreciate it if someone could tell me where i messed up. Thanks!
Sure There's something wrong here :

u=4x^2+1
u'=8x
du/8=dx
Basically, $u'=\frac{du}{dx}$.

So $du=8xdx \implies dx=\frac{du}{8x}$

The x in the denominator will simplify in the integral with the remaining x

$\int \frac{x}{(4x^2+1)^2} dx=\int \frac{x}{u^2} \frac{du}{8x}=\frac 18 \int \frac{1}{u^2} du=\frac 18 \int u^{-2} du$

Note that the integral $\int x^n dx=\frac{x^{n+1}}{n+1}$.

Here, n=-2 --> n=-1, not -3.

3. ## okay

Thanks. After I plug u back in I still get the wrong answer. If it's not to much of a hassle, could you finish the problem for me?

4. ## okay

I got that far. I then took the anti derivative of u^-2 and got -U^-1

I plugged u into that equation and got -((4x^2)+1)^(-1+2) = -((4x^2)+1)+c Then I multiply that by 1/8 which gives:

-((4x^2)+1)+c / 8

I must be messing up with the algebra part because I kept getting the same thing you posted.

My teacher gave -1/(32x^2+8) + c as the correct answer.

5. Originally Posted by PensFan10
I got that far. I then took the anti derivative of u^-2 and got -U^-1

I plugged u into that equation and got -((4x^2)+1)^(-1+2) = -((4x^2)+1)+c Then I multiply that by 1/8 which gives:

-((4x^2)+1)+c / 8

I must be messing up with the algebra part because I kept getting the same thing you posted.

My teacher gave -1/(32x^2+8) + c as the correct answer.
Hey ! Yes, you are messing up with the algebra (and the formula ?).

You have $\frac 18 \int u^{-2} du=\frac 18 (-u^{-1})=-\frac 18 \cdot \frac 1u=-\frac{1}{8(4x^2+1)}.$

Better ?

6. ## Awesome!

Yeah. That is much better. I should have just made it 1/u rather than trying to mess with the exponents. It all makes sense now