# Thread: Problem integrating after partial fractions

1. ## Problem integrating after partial fractions

(x + 301) / (x^3 + x+ 10)

Using synthetic division i factored the denominator to (x+2)(x^2 - 2x + 5)

After the partial fractions technique, A=23 B=-23 and C=93. i resulted with this:

int(23 / (x+3))dx + int(-23x / (x^2 -2x + 5))dx +int(93 / (x^2 -2x + 5))dx

i figured out the 1st part using the ln antiderivative. But i cant figure out how to integrate the 2nd 3rd term. Any help is great,
thanks

Chris

2. Hello, Chrs!

$\displaystyle \int\frac{x + 301}{x^3 + x + 10}\,dx$

Using synthetic division, i factored the denominator: .$\displaystyle (x+2)(x^2 - 2x + 5)$ . . . . Good!

After the partial fractions technique: .$\displaystyle A\,=\,23,\;B\,=\,-23,\;C\,=\,93$ . . . . Yes!

i had: .$\displaystyle 23\!\!\int\frac{dx}{x+23} \;-\;23\!\!\int\frac{x\,dx}{x^2 -2x + 5} \;+ \;93\!\!\int\frac{dx}{x^2 -2x + 5}$ . . . . Right!

i figured out the 1st part using the ln antiderivative.
But i cant figure out how to integrate the 2nd, 3rd terms.
Consider them as one fraction: .$\displaystyle \frac{-23x + 93}{x^2-2x+5}$

Now we'll perform some Olympic-level gymnastics on the numerator.

We have: .$\displaystyle -23x + 93 \;=\;-23x + 23 + 70$

Factor out -23: .$\displaystyle -23\left(x - 1 - \frac{70}{23}\right)$

Muliply by $\displaystyle \frac{2}{2}\!:\;\;-\frac{23}{2}\left(2x - 2 - \frac{140}{23}\right)$

And we have: .$\displaystyle -\frac{23}{2}(2x-2) + 70$

The integral becomes: .$\displaystyle \int\frac{-\frac{23}{2}(2x-2) + 70}{x^2-2x+5}\,dx$

. . $\displaystyle =\;-\frac{23}{2}\!\!\underbrace{\int\frac{2x-2}{x^2-2x+5}\,dx}_{\text{This is }\frac{dy}{u}} \;\;+\;\;70\!\!\underbrace{\int\frac{dx}{x^2-2x+5}}_{\text{Complete the square}}$

The last denominator is: .$\displaystyle x^2 -2x + 1 + 4 \:=\:(x-1)^2 + 4$

And the last integral is: .$\displaystyle 70\!\!\int\frac{dx}{(x-1)^2+4}$ . . . arctangent form!

I need a nap!
.

3. Hi Soroban, Thank you so much for the help. It really means alot.
I actually just have one more part that i really just want to confirm, im unsure if im done or not:
Using the previous findings (the 3 integrals), resolve with the improper integral
∫1-->∞ F(x)dx
That is 1 to infinity, pretty bad looking, i know

Anyways, for the 3 integrals(ln, ln and arctan) i got
∞, ∞, and π/2
So that would conclude that the integral goes to infinity?
Is there an answer ti find other than infinity? Sorry for the rediculous fonts.

4. Hello again, chrsr345!

The answer is finite . . . but it takes even more gymnastics!

We had: .$\displaystyle 23\int\frac{dx}{x+3} - \frac{23}{2}\int\frac{2x-2}{x^2-2x+5}\,dx + 70\int\frac{dx}{(x-1)^2 + 4}$

Integrate: .$\displaystyle 23\ln(x+3) - \frac{23}{2}\ln(x^2-2x+5) + 35\arctan\left(\frac{x-1}{2}\right)$

Factor out $\displaystyle \frac{23}{2}$ from the first two terms:

.$\displaystyle \frac{23}{2}\,\bigg[2\cdot\ln(x+3) - \ln(x^2-2x+5)\bigg] + 35\arctan\left(\frac{x-1}{2}\right)$

. . $\displaystyle = \;\frac{23}{2}\cdot\bigg[\ln(x+3)^2 - \ln(x^2-2x+5)\bigg] + 35\arctan\left(\frac{x-1}{2}\right)$

. . $\displaystyle = \;\frac{23}{2}\cdot\ln\left(\frac{x^2+6x+9}{x^2-2x+5}\right) + 35\arctan\left(\frac{x-1}{2}\right)$

Now we're supposed to evaluate from $\displaystyle 1$ to $\displaystyle \infty$.

If $\displaystyle x=1\!:\;\;\frac{23}{2}\ln\left(\frac{16}{4}\right) + 35\arctan(0) \;=\;\frac{23}{2}\ln(4) + 0 \;=\;\boxed{\frac{23}{2}\ln(4)}$

For "infinity": .$\displaystyle \lim_{x\to\infty}\bigg[\frac{23}{2}\ln\left(\frac{x^2+6x+9}{x^2-2x+5}\right) + 35\arctan\left(\frac{x-1}{2}\right)\bigg]$

. . $\displaystyle = \;\lim_{x\to\infty}\bigg[\frac{23}{2}\ln\left(\frac{1 + \frac{6}{x} + \frac{9}{x^2}}{1 - \frac{2}{x} + \frac{5}{x^2}}\right) + 35\arctan\left(\frac{x-1}{2}\right)\bigg]$

. . $\displaystyle = \;\frac{23}{2}\ln\left(\frac{1+0+0}{1-0+0}\right) + 35\arctan(\infty) \;= \;\frac{23}{2}\ln(1) + 35\left(\frac{\pi}{2}\right) \;=\;\boxed{\frac{35\pi}{2}}$

Therefore, the answer is: . $\displaystyle {\color{blue}\frac{35\pi}{2} - \frac{23}{2}\ln(4)}$