Integration by Parts: Solving Indefinite Integral (x+3)/(x^2+6x) dx

muddyjch
Messages
15
Reaction score
0

Homework Statement


Evaluate the indefinite integral.

∫(x + 3)/(x^2+6x) dx


Homework Equations


This is an online homework prob. that covers sections integration by parts and substitution in indefinite integrals. it looks to me that it fits into the formula ∫udv=uv-∫vdu if you change the original equation to (x+3)*(x^2+6x)^-1


The Attempt at a Solution


Doing this would give you u=x+3, dv=(x^2+6x)^-1, du=1, v=(-log(x+6)-log(x))/6

the final answer becomes (x+3)((-log(x+6)-log(x))/6)-((-(x+6)log(x+6)+xlog(x)-2x-6)/6)

this however is not correct answer. Am I using the wrong method?
 
Physics news on Phys.org
muddyjch said:
Am I using the wrong method?

You are missing an easier method. Take the derivative of the denominator and compare it to the expression in the numerator. Do you see an easy relation between the TWO? Now ask, what substitution makes sense?
 
I think you'd be better served trying substitution.
 
stevenb said:
You are missing an easier method. Take the derivative of the denominator and compare it to the expression in the numerator. Do you see an easy relation between the TWO? Now ask, what substitution makes sense?

Are you suggesting sub u=x^2+6x, du=2x+6 dx and so 1/2du=x+3 dx and the final answer should be 1/2 log (x^2+6x)? This is not the answer he wants because i believe he wants us to use a different method. Kind of confusing.
 
muddyjch said:
This is not the answer he wants because i believe he wants us to use a different method. Kind of confusing.

It is confusing. Did he say what method he wants you to use? You mentioned the problem came from an online HW problem directed at int. by parts and substitution. It seems substitution is the preferred and direct method. But, there is usually more than one way to skin a cat.
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
Back
Top