# Thread: Integration By Partial Fractions - Urgent Help Please

1. ## Integration By Partial Fractions - Urgent Help Please

integral(x/(x^2+4x+13),x,0,1)

I broke the bottom up to (x+2)^2 + 9, but now I"m stuck. any help would be very appreciated.

Thanks,
s3n4te

edit: i think i should now use inverse trig substitution.

2. Hi

$\frac{x}{x^2+4x+13} = \frac{\frac{1}{2}\2x+4)}{x^2+4x+13} - \frac{2}{(x+2)^2+9}" alt="\frac{x}{x^2+4x+13} = \frac{\frac{1}{2}\2x+4)}{x^2+4x+13} - \frac{2}{(x+2)^2+9}" />

The first part is integrable in $\frac{1}{2}\:ln|x^2+4x+13|$

The second part is integrable in Arctan after setting x+2=3t

3. How do you know how to split that up like that?

4. Originally Posted by s3n4te
How do you know how to split that up like that?
By experience

You make $\frac{u'}{u}$ appear eventually using a multiplicative constant - here $\frac{1}{2}$

The other part is then integrable with Arctan (this is true because the denominator has no real roots)