Mastering Integrals: Solving Tricky Equations with x and Quotient Rule

[zombieguy]

Homework Statement

How do I do these 2?

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

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

Homework Equations

The Attempt at a Solution

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

b) Tried substitution (because there's 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

 

[Dick]

There's no 'quotient rule' for integrals! Try partial fractions.

 

[darkmagic]

Factor first the denominator then use the partial fractions.

 

[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?

 

[Mark44]

Part b is not correct. After factoring you have
\int \frac{4}{(x + 2)^2}~dx

You can do this with an ordinary substitution. Your answer should NOT have a log in it!

 

[Dick]

If you mean (3/5)ln(x+3)+(2/5)ln(x-2), (use more parentheses!) then the first one looks ok. I don't like the looks of the second one. If you had shown your work, I might have been able to tell you where you went wrong.

 

[zombieguy]

OK, I've tried b) again but this time 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 nice if you could show me how to do it your way (using substitution) aswel

 

[Dick]

Let u=(x+2). du=dx. Then 4/(x+2)^2 dx=4/u^2 du=4*u^(-2) du. Use the power law integral u^n=u^(n+1)/(n+1).

 

[zombieguy]

Thank you all for your help