How do I do this integral? Partial Fractions?

[cerbeyx]

I tried to do partial fractions on this but it went very wrong and I'm not even sure that was the best method to use. How would you start this integral? $$\int \frac{1-2x}{(x-1)^2(x^2+1)}dx$$ Thanks for the help.

 

[BigBeachBanana]

I tried to do partial fractions on this but it went very wrong and I'm not even sure that was the best method to use. How would you start this integral? $$\int \frac{1-2x}{(x-1)^2(x^2+1)}dx$$ Thanks for the help.

I would start with partial fractions as well. Please share your attempt.

 

[Steven G]

Partial fractions.
If you want help, versus a complete solution, you need to show us your work so we can see where you are going wrong.

 

[limiTS]

Start by figuring out the values of A, B, C, and D:

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

 

[cerbeyx]

I don't know what to do from here. And thank you, limiTS, I set it up wrong the first time

 

[BigBeachBanana]

I don't know what to do from here. And thank you, limiTS, I set it up wrong the first time

I would expand the right-hand side and collect like terms. After that, equating the coefficients should give you a system of 4 equations and 4 unknowns.

(REF/RREF should help solve the linear system if you're familiar.)

 

[limiTS]

After you find that $A=-\frac{1}{2}$, update the partial fraction equation:
$$\frac{1-2x}{(x-1)^2(x^2+1)}=\frac{-\frac{1}{2}}{(x-1)^2}+\frac{B}{x-1}+\frac{Cx+D}{x^2+1}$$Move the known $A$ to the L.H.S.:
$$\begin{aligned} \frac{1-2x}{(x-1)^2(x^2+1)}+\frac{\frac{1}{2}}{(x-1)^2}&=\frac{B}{x-1}+\frac{Cx+D}{x^2+1}\\ \frac{1-2x}{(x-1)^2(x^2+1)}+\frac{\frac{1}{2}(x^2+1)}{(x-1)^2(x^2+1)}&=\\ \frac{1-2x+\frac{1}{2}x^2+\frac{1}{2}}{(x-1)^2(x^2+1)}&=\\ \frac{\frac{1}{2}x^2-2x+\frac{3}{2}}{(x-1)^2(x^2+1)}&=\\ \frac{\frac{1}{2}(x^2-4x+3)}{(x-1)^2(x^2+1)}&=\\ \frac{\frac{1}{2}\cancel{(x-1)}(x-3)}{(x-1)^{\cancel{2}}(x^2+1)}&=\\ \frac{\frac{1}{2}(x-3)}{(x-1)(x^2+1)}&=\frac{B}{x-1}+\frac{Cx+D}{x^2+1}\\ \end{aligned}$$
Expand this new equation (like you did before), and let $x=1$. You'll see that $B=-\frac{1}{2}$:

Move the known $B$ to the L.H.S.:

$$\begin{aligned} \frac{\frac{1}{2}(x-3)}{(x-1)(x^2+1)}&=\frac{-\frac{1}{2}}{x-1}+\frac{Cx+D}{x^2+1}\\ \frac{\frac{1}{2}(x-3)}{(x-1)(x^2+1)}+\frac{\frac{1}{2}}{x-1}&=\frac{Cx+D}{x^2+1}\\ \frac{\frac{1}{2}(x-3)}{(x-1)(x^2+1)}+\frac{\frac{1}{2}(x^2+1)}{(x-1)(x^2+1)}&=\\ \frac{\frac{1}{2}x-\frac{3}{2}}{(x-1)(x^2+1)}+\frac{\frac{1}{2}x^2+\frac{1}{2}}{(x-1)(x^2+1)}&=\\ \frac{\frac{1}{2}x^2+\frac{1}{2}x-1}{(x-1)(x^2+1)}&=\\ \frac{\cancel{(x-1)}(\frac{1}{2}x+1)}{\cancel{(x-1)}(x^2+1)}&=\\ \frac{\frac{1}{2}x+1}{x^2+1}&=\frac{Cx+D}{x^2+1}\\ \end{aligned}$$
Now it's obvious that $C=\frac{1}{2}$ and $D=1$.

Next, integrate!

$$\int\left[\frac{-\frac{1}{2}}{(x-1)^2}+\frac{-\frac{1}{2}}{x-1}+\frac{\frac{1}{2}x+1}{x^2+1}\right]\text{d}x$$

 

[skeeter]

$\dfrac{1-2x}{(x-1)^2(x^2+1)} = \dfrac{A}{(x-1)^2} + \dfrac{B}{x-1} + \dfrac{Cx+D}{x^2+1}$

$1-2x = A(x^2+1)+B(x-1)(x^2+1)+(Cx+D)(x-1)^2$

$1-2x = (B+C)x^3 + (A-B-2C+D)x^2 + (B+C-2D)x + (A-B+D)$

equate coefficients ...

$B+C=0$
$A-B-2C+D = 0$
$B+C-2D = -2$
$A-B+D = 1$

solving the system yields ...

$$A= -\dfrac{1}{2} \text{ ; } B = -\dfrac{1}{2} \text{ ; } C = \dfrac{1}{2} \text{ ; } D = 1$$