OKComputer
New member
- Joined
- Sep 13, 2014
- Messages
- 3
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!
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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