Question

find the indefinite integral

Answer 1

\[\int\limits \frac{2x+3}{x^2+4x+13} dx\]

Answer 2

Um why did you post that, and why did someone give you a medal?

Answer 3

so imo we can do this 2 ways.

Answer 4

I'm taking care of that right now

Answer 5

Thanks turning, you're the best :D.

Answer 6

brb

Answer 7

We could split this integral up into \[\int\limits \frac{2x}{x^2+4x+13} + \int\limits \frac{3}{x^2+4x+13} dx\]

Answer 8

but I think we should use the normal num trying to do this with "linear fractions?"

Answer 9

ok I'm back
dang trolls...
so one way to do this is with partial fractions, and I'm sure another will come to me in a sec

Answer 10

The only way to get this into partial fractions would be (x+2)^2 + 9?

Answer 11

And yeah man WTF was that LOL... He comes in with a copy/paste, and his friend gives him a medal....?

Answer 12

I think he made a duplicate account and gave himself a medal

Answer 13

Idiots....

Answer 14

actually I was wrong about the partial fractions idea... I can only see one way to do this right now

notice that the numerator is almost the derivative of the denominator...

Answer 15

so we can rewrite this as\[\int\frac{2x+3}{x^2+4x+13}dx=\int\frac{2x+4}{x^2+4x+13}-\frac1{x^2+4x+13}dx\]starting to see where this is going?

Answer 16

\[u=x^2+4x+13\]will handle the first integral
for the second you need to be a bit more clever and complete the square in the denominator

Answer 17

okay,sorry I was dealing with Verizon Wireless.

Oh hmm that's really interesting what you did there :D

Answer 18

u = x^2+4x+4
du = 2x+4dx

Answer 19

\[\int\limits \frac{du}{u} -\int\limits \frac{1}{x^2+4x+13} dx\]

Answer 20

then for the 2nd oneI could I do

(x+2)^2 + 9

u = x+2
du = dx

a = 3

Answer 21

\[\int\limits \frac{du}{u} -\int\limits \frac{du}{u^2+a^2} dx\]

Answer 22

\[\ln|u| - \frac{1}{a}\arctan(\frac{u}{a}) + c\]

Answer 23

\[\ln|x^2+4x+4| - \frac{1}{3}\arctan(\frac{x+2}{3}) + c\]

Answer 24

err ln|x^2+4x+13|

Answer 25

yep :)

Answer 26

:D

Answer 27

I don't see another way to do it yet though...

Answer 28

TO THE NEXT QUESTIONS!

I'm starting to get good at these :).

Answer 29

Yeah the partial fraction was Idk...

Answer 30

you sure are!

Answer 31

:)

Answer 32

NEXT QUESTION!