# Thread: Stuck on an integral

1. ## Stuck on an integral

I'm have a bit of trouble with this integral
$\int\frac{dx}{(x+1)(x^2 +4)}$
I'm pretty sure I have to use parts on it but it's been awhile, any help would be appreciated.

2. Originally Posted by doughnuts
I'm have a bit of trouble with this integral
$\int\frac{dx}{(x+1)(x^2 +4)}$
I'm pretty sure I have to use parts on it but it's been awhile, any help would be appreciated.
PFD (Partial Fractions Decomposition)

$\frac{1}{(x+1)(x^2+4)}=\frac{A}{x+1}+\frac{Bx+C}{x ^2+4}$

Multiplying through by the LCD we get

$1=A(x^2+4)+(Bx+C)(x+1)$

Now if we let $x=-1$

we get

$1=A(1+5)+(B(-1)+C)(-1+1)\Rightarrow{A=\frac{1}{5}}$

So now since the other expression has no real roots we must expand

$1=Ax^2+4A+Bx^2+Bx+Cx+C$

Equating coefficients we get

$A+B=0\Rightarrow{B=\frac{-1}{5}}$

and

$4A+C=1\Rightarrow{C=\frac{1}{5}}$

This gives us

$\frac{1}{(x+1)(x^2+4)}=\frac{1}{5}\bigg[\frac{1}{x+1}-\frac{x}{x^2+4}+\frac{1}{x^2+4}\bigg]$

Which implies that

$\int\frac{dx}{(x+1)(x^2+4)}=\frac{1}{5}\int\bigg[\frac{1}{x+1}-\frac{x}{x^2+4}+\frac{1}{x^2+4}\bigg]$ $=\frac{1}{5}\bigg[\ln|x+1|-\frac{1}{2}\ln(x^2+4)+\frac{1}{2}\arctan\left(\fra c{x}{2}\right)\bigg]+C$

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Just out of curiosity how were you planning to use parts

$u=\frac{1}{x+1}$
$dv=\frac{1}{x^2+4}$?

3. Originally Posted by Mathstud28
Just out of curiosity how were you planning to use parts

$u=\frac{1}{x+1}$
$dv=\frac{1}{x^2+4}$?
Yeah that's what I was trying to use, though now that I see it it looks like partial fractions would be the best way to go, thanks for the speedy response.

4. Originally Posted by doughnuts
Yeah that's what I was trying to use, though now that I see it it looks like partial fractions would be the best way to go, thanks for the speedy response.
Anytime

Just remember generally if you are given

$\int\frac{N(x)}{D(x)}dx$

Where $N(x)$ and $D(x)$ are polynomials, and $D(x)$ is reducible try PFD

5. Originally Posted by Mathstud28
Anytime

Just remember generally if you are given

$\int\frac{N(x)}{D(x)}dx$

Where $N(x)$ and $D(x)$ are polynomials, and $D(x)$ is reducible try PFD
Check first that $deg(N), where "deg" = degree

6. I hate to bring this thread back from the dead, but after I had this question answered I went on to try and find out if the integral was convergent or divergent from -1 to 6. My answer seemed to be way smaller then it should of been, perhaps I'm doing this wrong.
I know that in the original equation that X = -1 was a vertical asymptote so I'd should set
$\lim_{t \to -1} \int_t^{6}\frac{dx}{(x+1)(x^2 +4)}$
then
$f(x) = \frac{1}{5}\bigg[\ln|7|-\frac{1}{2}\ln(40)+\frac{1}{2}\arctan\left(3\right )\bigg]$
Subtract
$\frac{1}{5}\bigg[\ln|0|-\frac{1}{2}\ln(5)+\frac{1}{2}\arctan\left(\frac{-1}{2}\right)\bigg]$ should equal what the area under the curve is.
any help would be appreciated.

7. Originally Posted by doughnuts
I hate to bring this thread back from the dead, but after I had this question answered I went on to try and find out if the integral was convergent or divergent from -1 to 6. My answer seemed to be way smaller then it should of been, perhaps I'm doing this wrong.
I know that in the original equation that X = -1 was a vertical asymptote so I'd should set
$\lim_{t \to -1} \int_t^{6}\frac{dx}{(x+1)(x^2 +4)}$
then
$f(x) = \frac{1}{5}\bigg[\ln|7|-\frac{1}{2}\ln(40)+\frac{1}{2}\arctan\left(3\right )\bigg]$
Subtract
$\frac{1}{5}\bigg[\ln|0|-\frac{1}{2}\ln(5)+\frac{1}{2}\arctan\left(\frac{-1}{2}\right)\bigg]$ should equal what the area under the curve is.
any help would be appreciated.

Ln(0) doesn't exist, the limit as x-> 0 of ln(x) is negative infinity. Hence in the latter portion you would be subtracting this value which would cancel out the negative and make it positive infinity. Hence the improper integral diverges and the area is infinite.

Other than not pointing that out, you're right.