Question

large int_{}^{} \frac{ 2x }{ x^2+6x+13 }

Could someone help please? :3
I've learned logs and inverse trig so far :)

Answer 1

Latex is broken, but seems you just need to get the derivative of the bottom?

Answer 2

\[\large \int\limits_{}^{} \frac{ 2x }{ x^2+6x+13 } \]

Answer 3

derivative of bottom is 2x + 6 ?

Answer 4

Yeah I'm thinking you have to do a u sub with x^2+6x+13 :d

Answer 5

Try it out, and see what you get

Answer 6

(du+6)/u ? lol 0.o

Answer 7

maybe split it into two integrals first

Answer 8

\(\large \int \frac{2x}{x^2+6x+13} = \int \frac{2x+6-6}{x^2+6x+13} \)


\(\large = \int \frac{2x+6}{x^2+6x+13} + \int \frac{-6}{x^2+6x+13}\)

Answer 9

you could try u-sub for first integral,
and a trig-sub for last integral

Answer 10

The derivative of the denominator is 2x+6, which you unfortately do not have in the numerator. Follow the advice from ganeshie8, or, alternatively, "complete the square" of the denominator.

Answer 11

Yeah, I just did complete the square, and got the right answer :d

Answer 12

Cool. guide Jigglypuff towards doing this herself, if she asks you to.

Answer 13

I understand that 1/(x^2+6x+13) -> (1/2)arctan((x+3)/2)
but idk what the -6 in the numerator does to that...

Answer 14

First of all, ganeshie8's approach WILL work. I'll leave it to him/her to explain where those 6's in the numerator came from and why they're helpful.

Answer 15

\(\large \int c f(x) dx = c\int f(x) dx\)

Answer 16

just pull out the -6 out of integral

Answer 17

\(\large \int \frac{2x}{x^2+6x+13} = \int \frac{2x+6-6}{x^2+6x+13} \)

\(\large = \int \frac{2x+6}{x^2+6x+13} + \int \frac{-6}{x^2+6x+13}\)

\(\large = \int \frac{2x+6}{x^2+6x+13} -6 \int \frac{1}{x^2+6x+13}\)

Answer 18

Oh right ^_^
ok, I'll try that, thank you :)

Answer 19

np :) did u complete the square for second integral and did trig-sub ?

Answer 20

yep :)
I got it, thank you! :D

Answer 21

okay nice :)

Answer 22

numerator : x = 2u-3

Answer 23

@ganeshie8 good point!