Please, help me with this integral.

OKComputer

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Hello to you all, I'm new to integral calculus and I've been having some troubles especially with that kind of problems. If you can explain me how to solve it correctly and which method you applied then I'd be very grateful.

Problem:. . .\(\displaystyle \displaystyle{\int\, \frac{(3x\, -\, 5)\, dx}{x^2\, +\, 6x\, +\, 18}}\)

So Far Now:

\(\displaystyle \displaystyle{\int\, \frac{(3x\, -\, 5)\, dx}{x^2\, +\, 6x\, +\, 18}\, =\, \int\, \frac{(3x\, -\, 5)\, dx}{(x^2\, +\, 6x\, +\, 9)\, +\, 9}\, =\, \int\, \frac{(3x\, -\, 5)\, dx}{(x\, +\, 3)^2\, +\, (3)^2}}\)

\(\displaystyle u\, =\, x\, +\, 3\, \rightarrow\, u\, -\, 3\, =\, x\, \rightarrow\, du\, =\, dx\)

\(\displaystyle \displaystyle{\int\, \frac{3(u\, -\, 3)\, -\, 5}{u^2\, +\, 3^2}\, du\, \rightarrow\, \int\, \frac{3u\, -\, 9\, -\, 5}{u^2\, +\, 3^2}\, du\, \rightarrow\, 3\,\int\, \frac{u\, du}{u^2\, +\, 3^2}\, -\, 14\, \int\, \frac{du}{u^2\, +\, 3^2}}\)

I know that what I did may be completely wrong but it's because I really don't know how to solve it, in this case I'm sorry about my mistakes.

Thanks in advance for your help!
 
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\(\displaystyle \displaystyle{...\rightarrow\, 3\,\int\, \frac{u\, du}{u^2\, +\, 3^2}\, -\, 14\, \int\, \frac{du}{u^2\, +\, 3^2}}\)
The first integral looks like a u-sub would work (say, "let v = u^2 + 9, so dv = 2u du, so (1/2) dv = u du") and the second integral would work with an inverse-trig sub (using the inverse tangent, I think). ;)
 
Hello to you all, I'm new to integral calculus and I've been having some troubles especially with that kind of problems. If you can explain me how to solve it correctly and which method you applied then I'd be very grateful.

Problem:. . .\(\displaystyle \displaystyle{\int\, \frac{(3x\, -\, 5)\, dx}{x^2\, +\, 6x\, +\, 18}}\) ...

I would go about it a slightly different way. Notice that with little manipulation
\(\displaystyle \int\frac{(3x\, -\, 5)\, dx}{x^2\, +\, 6x\, +\, 18}=\frac{3}{2}\int\frac{(2x+6)dx}{x^2+6x+18}-14\int\frac{dx}{x^2+6x+18}\)
The first integral is for a perfect differential (and is equivalent to yours if you multiply and divide yours by 2) and the second leads to what you have with the substitution. Of course the two approaches are equivalent but just though I would do it a little differently.
 
Thank you very much stapel and Ishuda, now I know how to solve that kind of exercises! Thanks for your time helping me with this question!
 
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