# Thread: Help with 2 integrals

1. ## Help with 2 integrals

How do I do these 2?

a) x/(x^2+x-6)

Tried quotient rule it gave me -x^2 - 6/(x^2+x+6)^2, I don't think it's right

b) 4/(x^2+4x+4)

Tried substitution (because theres another similar question between these two on the paper and that was the way to go for it) Don't think my answers worth mentioning

2. Originally Posted by zombieguy
How do I do these 2?

a) x/(x^2+x-6)

Tried quotient rule it gave me -x^2 - 6/(x^2+x+6)^2, I don't think it's right

b) 4/(x^2+4x+4)

Tried substitution (because theres another similar question between these two on the paper and that was the way to go for it) Don't think my answers worth mentioning
a) $\frac{x}{x^2+x-6} = \frac{x}{(x+3)(x-2)}$ then try partial fractions

b) $\frac{4}{x^2+4x+4} = \frac{4}{(x+2)^2}$ try a $u$ substitution.

3. Ok, Thanks

I've now got:

a) 3/5ln(x+3)+2/5ln(x-2)

b) 2/(x+2)ln(x+2)^2

Is that right?

4. Originally Posted by zombieguy
Ok, Thanks

I've now got:

a) 3/5ln(x+3)+2/5ln(x-2)

b) 2/(x+2)ln(x+2)^2

Is that right?
The first is (but don't forget the absolute bars), the second isn't. It should be

$\frac{-4}{x+2} + c$.

5. OK, I was able to get your answer for b) but by using:

u.v-(integral of)(v.du/dx)

u=4
du/dx=0 (That gets rid of the integral part)

v=-(x+2)^-1
dv/dx= (x+2)^-2

so: 4.-(x+2)^-1

It would be nicee if you could show me how to do it your way (using substitution) aswel

6. $\int{\frac{4}{x^2 + 4x + 4}\,dx} = \int{\frac{4}{(x + 2)^2}\,dx}$.

Let $u = x + 2$ so that $\frac{du}{dx} = 1$.

So the integral becomes

$\int{\frac{4}{u^2}\,\frac{du}{dx}\,dx}$

$= \int{4u^{-2}\,du}$

$= -4u^{-1} + C$

$= -\frac{4}{x + 2} + C$.

7. Thank you both for your help