# Thread: Trouble with Partial Fraction Integration

1. ## Trouble with Partial Fraction Integration

Hello, I was hoping someone could pick off where I went wrong on these two questions. As far as I can tell I have the right answers but the program I am plugging the answers into says nay. Not sure on how to do the integral symbol on this forum so just assume it is there.

1) Integrate by partial fractions $-2x^2-4x+8/(X+1)(X^2+4)$

Using the partial fractions I was able to change it to $2/(x+1)-4/(X^2+4)$ which is so far correct according to the program (in bx+c on the top of the second fraction the c is 0)

So I am down to integrating $2/(x+1)$ and $-4/(x^2+4)$ . Using the natural log rule the first I figure is 2ln(x+1) and the second I solved using $u=x^2+4$ and getting 1/2du=x. The 1/2 comming outside of the integral it leaves 1/2 the integral of du/u, which is 1/2ln(u), which then gives you -4(1/2)ln(x^2+4) as the second integral.

So I figured $2(ln(x+1))-2ln(x^2+4)$ should be the answer, but it comes back as wrong and I am running out of attempts with no clue where I am wrong on this.

Second problem.

Integral of $-33e^x-126/e^(2x)+9e^x+18$ A hint is given to use u=e^x which I did getting -33u-126/u^2+9u+18, factoring the bottom leaves (u+3) and (u+6) for the partial fractions. I got -9/(u+3) and -24/(u+6) for the two partial fractions to solve. I ended up with -9ln(e^x+3)-24ln(e^x+6) when intrgrating and putting back in e^x for u, which is wrong.

Can anyone help me with where I am going wrong on either or both of these questions? I am guessing it might be the same problem on each, missing something I am supposed to do.

2. Originally Posted by LostInCalculus
Hello, I was hoping someone could pick off where I went wrong on these two questions. As far as I can tell I have the right answers but the program I am plugging the answers into says nay. Not sure on how to do the integral symbol on this forum so just assume it is there.

1) Integrate by partial fractions $-2x^2-4x+8/(X+1)(X^2+4)$

Using the partial fractions I was able to change it to $2/(x+1)-4/(X^2+4)$ which is so far correct according to the program (in bx+c on the top of the second fraction the c is 0)

So I am down to integrating $2/(x+1)$ and $-4/(x^2+4)$ . Using the natural log rule the first I figure is 2ln(x+1) and the second I solved using $u=x^2+4$ and getting 1/2du=x. The 1/2 comming outside of the integral it leaves 1/2 the integral of du/u, which is 1/2ln(u), which then gives you -4(1/2)ln(x^2+4) as the second integral.

Mr F says: NO! First of all, it's a standard form: $\int \frac{a}{x^2 + a^2} = \tan^{-1}\left( \frac{x}{a} \right)$. As to why and where your substitution goes wrong, see below.

So I figured $2(ln(x+1))-2ln(x^2+4)$ should be the answer, but it comes back as wrong and I am running out of attempts with no clue where I am wrong on this.

[snip]
$u = x^2 + 4 \Rightarrow dx = \frac{du}{{\color{red}2x}}$. So you have $\int \frac{1}{u} \frac{du}{{\color{red}2x}}$, which is no good because there's still an x there ..... Replace the x with $\sqrt{u - 4}$ and your troubles are worse .....

$\int \frac{4}{x^2 + 4} \, dx = 2\int \frac{2}{x^2 + 2^2} \, dx = 2 \tan^{-1} \left( \frac{x}{2} \right)$.

3. Originally Posted by LostInCalculus
[snip]Second problem.

Integral of $-33e^x-126/e^(2x)+9e^x+18$ A hint is given to use u=e^x which I did getting -33u-126/u^2+9u+18, factoring the bottom leaves (u+3) and (u+6) for the partial fractions. I got -9/(u+3) and -24/(u+6) for the two partial fractions to solve. I ended up with -9ln(e^x+3)-24ln(e^x+6) when intrgrating and putting back in e^x for u, which is wrong.
[snip]
You have to substitute for the dx as well, old chap!

$u = e^x \Rightarrow \frac{du}{dx} = e^x \Rightarrow dx = \frac{du}{e^x} = \frac{du}{u}$. So the integral you got is missing a factor of u in its denominator ...... Back to the partial fraction drawing board ....

4. Ack! I wrote that down a tiny bit wrong, it is actually -4x/(X^2+4), given that -4 was b but the x is still in the bx+c which changes stuff abit. I ran it through Wolframs integrator and it gives me the same -2ln(x^2+4) answer that I got the long way that comes up as wrong. I got the answer for the second question finally though, your catching that e^x thing helped huge.

Thanks for the help Mr. F