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? [math]\int \frac{1-2x}{(x-1)^2(x^2+1)}dx[/math] 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? [math]\int \frac{1-2x}{(x-1)^2(x^2+1)}dx[/math] 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:

[imath]\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}[/imath]

 

[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 [imath]A=-\frac{1}{2}[/imath], update the partial fraction equation:
[math]\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}[/math]Move the known [imath]A[/imath] to the L.H.S.:
[math]\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}[/math]
Expand this new equation (like you did before), and let [imath]x=1[/imath]. You'll see that [imath]B=-\frac{1}{2}[/imath]:

Move the known [imath]B[/imath] to the L.H.S.:

[math]\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}[/math]
Now it's obvious that [imath]C=\frac{1}{2}[/imath] and [imath]D=1[/imath].

Next, integrate!

[math]\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[/math]

 

[skeeter]

[imath]\dfrac{1-2x}{(x-1)^2(x^2+1)} = \dfrac{A}{(x-1)^2} + \dfrac{B}{x-1} + \dfrac{Cx+D}{x^2+1}[/imath]

[imath]1-2x = A(x^2+1)+B(x-1)(x^2+1)+(Cx+D)(x-1)^2[/imath]

[imath]1-2x = (B+C)x^3 + (A-B-2C+D)x^2 + (B+C-2D)x + (A-B+D)[/imath]

equate coefficients ...

[imath]B+C=0[/imath]
[imath]A-B-2C+D = 0[/imath]
[imath]B+C-2D = -2[/imath]
[imath]A-B+D = 1[/imath]

solving the system yields ...

[math]A= -\dfrac{1}{2} \text{ ; } B = -\dfrac{1}{2} \text{ ; } C = \dfrac{1}{2} \text{ ; } D = 1[/math]