Question

Calc II: Integration, by parts?
Question attached.

Answer 1

I assume you need help integrating\[\int\frac{2x+2}{x^2+1}dx?\]

Answer 2

Yeah, I'm not sure if I'm supposed to split it up again?

Answer 3

Id split it up into 2x/(x^2 + 1) + 2/(x^2 + 1)

Answer 4

You could split it, like this:\[\int\frac{2x}{x^2+1}dx+\int\frac{2}{x^2+1}dx.\]

Answer 5

Ok, and one other thing, is this legal?

\[\int\limits_{}^{} 2/(x-1) = 2 \int\limits_{}^{} 1/(x-1)\]

Answer 6

Yes.

Answer 7

Yep

Answer 8

I didn't think that you could bring the 2 out with it being the numerator, divided by the denominator

Answer 9

You can take anything out, except a variable.

Answer 10

As long as it's not a variable being integrated and it's factorisable, you can take it out.

Answer 11

Mk, so that works for the numerator, but not the denominator, right?

Answer 12

Works for the denominator.

Answer 13

\[\int\frac{1}{2x}dx=\frac{1}{2}\int\frac{1}{x}dx\]

Answer 14

You're fast..

Answer 15

^^

Answer 16

Ahh, so I can change that from 2/(x-1) to \[-2\int\limits_{}^{}1/x\]

Answer 17

Nope. \[\int\limits{ 2 \over x-1} dx = 2 \int\limits {1 \over x-1}\]

Answer 18

dx

Answer 19

but didn't you say we could take that -1 from the denominator?

Answer 20

It has to be factorisable.

Answer 21

Think of it as normal factorization. Just instead of the bracket, there's a cool looking integral sign

Answer 22

Ahh, I see now

Answer 23

Thanks!

Answer 24

! last thing, this is what my friend emailed me. On the second integral did he just forget the 2x in the numerator?

Answer 25

And by friend I mean professor laughing that I got this one wrong,

Answer 26

Your friend's work is correct.

Answer 27

professor*

Answer 28

Mk, just checking,
Thanks again!