1. ## Integration Help

I need to integrate 1/(16x^3-4x^2+4x-1)

I first split the denominator into (4x-1)(4x^2+1) and used partial fraction decomposition to get:

the integral of (4/5)/(4x-1)+((-4/5)x-(1/5))/(4x^2+1)

I know how to solve the first segment of the integral, but I don't know how to integrate ((-4/5)x-(1/5))/(4x^2+1)

2. Originally Posted by Simon777
I need to integrate 1/(16x^3-4x^2+4x-1)

I first split the denominator into (4x-1)(4x^2+1) and used partial fraction decomposition to get:

the integral of (4/5)/(4x-1)+((-4/5)x-(1/5))/(4x^2+1)

I know how to solve the first segment of the integral, but I don't know how to integrate ((-4/5)x-(1/5))/(4x^2+1)
Try splitting it into two rational expressions - the first with (-4/5)x in the numerator and the second with -(1/5) in the numerator.
The first one will involve ln(4x^2+1),
the second will need substitution.

3. So you have:

$\int \frac{4}{5(4x-1)}-\frac{4x+1}{5(x^{2}+1)}dx \Rightarrow$

$\int \frac{4}{5(4x-1)}-\frac{4x}{5(x^{2}+1)}+\frac{1}{5(x^{2}+1)}dx$

$u = 4x-1; du = 4dx$

$w = x^2+1; dw = 2xdx$

$\int \frac{du}{5u}-\frac{2dw}{5w}+\frac{1}{5(x^{2}+1)} \Rightarrow$

$\frac{ln|4x-1|-ln(x^2+1)^{2}+tan^{-1}(x)}{5}$

4. Originally Posted by ANDS!
So you have:

$\int \frac{4}{5(4x-1)}-\frac{4x+1}{5(x^{2}+1)}dx \Rightarrow$
With everything after the subtraction sign, why is it just (x^2+1)? Shouldn't it be (4x^2+1)? That was what the original problem was split into.

5. Originally Posted by Debsta
Try splitting it into two rational expressions - the first with (-4/5)x in the numerator and the second with -(1/5) in the numerator.
The first one will involve ln(4x^2+1),
the second will need substitution.
Thank you, I got a little farther with that advise, but what is the integral of:

1/(4x^2+1)

I can't do u substitution and I can't convert this into tan^-1 because of the 4.

6. Originally Posted by Simon777
Thank you, I got a little farther with that advise, but what is the integral of:

1/(4x^2+1)

I can't do u substitution and I can't convert this into tan^-1 because of the 4.
edit: mistake. refer to skeeter's post

7. Originally Posted by Simon777
Thank you, I got a little farther with that advise, but what is the integral of:

1/(4x^2+1)

I can't do u substitution and I can't convert this into tan^-1 because of the 4.
$\frac{1}{4x^2+1} = \frac{1}{(2x)^2+1} = \frac{1}{2} \cdot \frac{2}{(2x)^2+1}$

$\int \frac{1}{2} \cdot \frac{2}{(2x)^2+1} \, dx$

$\frac{1}{2}\arctan(2x) + C$

8. Originally Posted by Simon777
With everything after the subtraction sign, why is it just (x^2+1)? Shouldn't it be (4x^2+1)? That was what the original problem was split into.
It is. Doesn't change using trig sub though. Even if it wasn't a perfect square.

9. Thank you all so much for the help. Can someone please check my answer?

ln|((4x-1)^(1/5)) / ((4x^2+1)^(1/10))| - (1/10)arctan(2x) + C

10. Originally Posted by Simon777
Thank you all so much for the help. Can someone please check my answer?

ln|((4x-1)^(1/5)) / ((4x^2+1)^(1/10))| - (1/10)arctan(2x) + C
Yes that looks right. Just one thing, because of the yukky exponents in the first term, I would have written the first term as:
(1/5) ln|(4x-1)/sqr(4x^2+1)|

11. Originally Posted by Debsta
Yes that looks right. Just one thing, because of the yukky exponents in the first term, I would have written the first term as:
(1/5) ln|(4x-1)/sqr(4x^2+1)|
Thanks for taking the time to check that.