1. ## partial fractions

I understand how to do these, but I am stuck on how to solve for these 4 variables i got in these equations:
A + C = 2
B + D = 0
A + C = 0
B + D = 0

..just incase i messed up..my original equation is
integral of (2x^3)/((x^2 + 1))^2
So i did 2x^3)/(x^2 + 1)(x^2 + 1) = (Ax + B)/(x^2 + 1) +
(Cx + D)/(X^2 + 1)

I then got Ax^3 + Ax + Bx^2 + B + Cx^3 + Cx + Dx^2 + D
Then I got those 4 equations above...I am not seeing how to solve for the variables right now..maybe a brain fart.

2. $\displaystyle \frac{2x^3}{(x^2+1)^2} = \frac{Ax+b}{x^2+1} + \frac{Cx+d}{(x^2+1)^2}$. In the second term you forgot to square it.

3. Doing the PF thing:

$\displaystyle \frac{Ax+B}{x^{2}+1}+\frac{Cx+D}{(x^{2}+1)^{2}}=2x ^{3}$

Multiply by the LCD $\displaystyle (x^{2}+1)^{2}$, expand out and we get:

$\displaystyle Ax^{3}+Bx^{2}+Ax+B+Cx+D=2x^{3}$

Equate coefficients:

A=2
B=0
A+C=0
B+D=0

Two of the vaiables are already solved, so

solve this system and we can see A=2, B=0, C=-2, D=0

So, you have:

$\displaystyle \frac{2x}{x^{2}+1}-\frac{2x}{(x^{2}+1)^{2}}$

4. Originally Posted by tukeywilliams
$\displaystyle \frac{2x^3}{(x^2+1)^2} = \frac{Ax+b}{x^2+1} + \frac{Cx+d}{(x^2+1)^2}$. In the second term you forgot to square it.

I am getting equations that make no sense now..

Ax^3 + Ax + Bx^2 + B + Cx^5 + 2Cx^3 + Cx + Dx^4 + 2Dx^2 + D
so..
for the x^5 terms: C = 0
x^4 terms: D = 0
X^3 terms: A + 2C = 2 A = 2 since C = 0
X^2 terms: B + 2D = 0 B = 0 since D = 0
X^1 terms: A + C = 0 (impossible since A = 2)
X^0 terms: B + D = 0

5. Originally Posted by davecs77
I am getting equations that make no sense now..

Ax^3 + Ax + Bx^2 + B + Cx^5 + 2Cx^3 + Cx + Dx^4 + 2Dx^2 + D
so..
for the x^5 terms: C = 0
x^4 terms: D = 0
X^3 terms: A + 2C = 2 A = 2 since C = 0
X^2 terms: B + 2D = 0 B = 0 since D = 0
X^1 terms: A + C = 0 (impossible since A = 2)<---what if C=-2
X^0 terms: B + D = 0
I think you have an oversight.

6. Originally Posted by davecs77
I understand how to do these, but I am stuck on
integral of (2x^3)/((x^2 + 1))^2
Just in case, there's no necessity to apply partial fractions.

$\displaystyle \int\frac{2x^3}{(x^2+1)^2}~dx$

Set $\displaystyle u=x^2+1\implies{du}=2x~dx$, therefore

\displaystyle \begin{aligned} \int {\frac{{2x^3 }} {{(x^2 + 1)^2 }}~dx} &= \int {\frac{{u - 1}} {{u^2 }}~du}\\ &= \ln \left| u \right| + \frac{1} {u} + k\\ &= {\color{blue}\ln \left(x^2 + 1\right) + \frac{1} {{x^2 + 1}} + k},~~k\in\mathbb{R}~\blacksquare \end{aligned}