Question

Integrating, what method should I use?

Answer 1

\[\int\limits_{}^{}\frac{ x+1 }{ 9x^2+6x+5 }dx\]

Answer 2

So I was thinking partial fraction decomposition? But I can't factor out the bottom nicely.

Answer 3

nah completing the square

Answer 4

ou. okay let me try.

Answer 5

nope the factors will contain imaginary numbers

Answer 6

how so?

Answer 7

So it would be
(3x+1)^2+4 ?

Answer 8

oh yeah right

Answer 9

So then we could rewrite it
\[\int\limits_{}^{}\frac{ x+1 }{ (3x+1)^2+2^2 }\]

Answer 10

Or I guess we don't need to write the 4 like that..hm

Answer 11

then what?

Answer 12

sorry was helping others

Answer 13

No worries :)

Answer 14

so now you know 1/(a^2+x^2)= tanx right

Answer 15

Yes
well..isn't it the square root of a^2+x^2 =atanx ?

Answer 16

nop check your identities again

Answer 17

are you referring to the integration formula?

Answer 18

\[\int\limits_{}^{}\frac{ 1 }{ a^2+x^2 }=\frac{ 1 }{ a }\arctan \frac{ x }{ a }+C\]

Answer 19

@zepdrix :>

Answer 20

yeah break in to two parts

Answer 21

What two parts?

Answer 22

\[\int\limits_{}^{}\frac{ 1 }{(3x+1)^2+2^2 }*x+1 dx\] this?

Answer 23

no

Answer 24

\[\int\limits_{}^{}\frac{ x+1 }{ (3x+1)^2+2^2 }=\int\frac{x}{(3x+1)^2+4}dx+\frac{1}{(3x+1)^2+4}dx\]

Answer 25

second will produce arcan, first is a simple u-sub

Answer 26

i missed an integral sign there, but you get the idea i hope

Answer 27

Yes, I do :)
is the u=3x+1 ? then the du comes out to 3 which doesn't help much?

Answer 28

\[\int\limits_{}^{}\frac{ x+1 }{ (3x+1)^2+2^2 }=\int\frac{x}{(3x+1)^2+4}dx+\int\frac{1}{(3x+1)^2+4}dx\]

Answer 29

oh yeah you are right i screwed up

Answer 30

well i just sort of screwed up
starting with \[\int\limits_{}^{}\frac{ x+1 }{ (3x+1)^2+2^2 }\] put \(u=3x+1\) now, then break it in to two parts

Answer 31

...ok xD

Answer 32

it is going to be ugly keeping track of all the constants, so be careful
good luck

Answer 33

\[3[\int\limits_{}^{}\frac{ x }{ u^2+4 }du+\int\limits_{}^{}\frac{ 1 }{ u^2+4 }du]\]??

Answer 34

\(\color{#000000}{ \displaystyle \int \frac{x}{(3x+1)^2+4} dx +\int \frac{1}{(3x+1)^2+4} dx }\)

For the first integral,
\(\color{#000000}{ \displaystyle u=3x+1 \quad \Longrightarrow \quad x=(u-1)/3 }\)
\(\color{#000000}{ \displaystyle du=3~dx \quad \Longrightarrow \quad du/3=dx }\)

\(\color{#000000}{ \displaystyle \int \frac{(u-1)/3}{u^2+4} (du/3) }\)

\(\color{#000000}{ \displaystyle (1/9)\int \frac{u-1}{u^2+4} du }\)

\(\color{#000000}{ \displaystyle (1/18)\int \frac{2u}{u^2+4} du-(1/9)\int \frac{1}{u^2+4} du }\)

and use the integral for arctangent.


For the second integral,
\(\color{#000000}{ \displaystyle \int \frac{1}{(3x+1)^2+2^2} dx }\)
use the integral for arctangent.

Answer 35

I believe there was an easier way to begin, but won't mess up once we have gone this far.

Answer 36

^ ah well this is just practice for an exam. Since speed counts in an exam, perhaps it is worth it to go through the easier way? :) But We can finish this one out as well

Answer 37

Well, your teachers are not bastards, are they?

Answer 38

Although, yeah, perhaps they might give you something as dirty, but they would then give you some time to work it.

I think I know an easier way to do this integral.

Answer 39

Weeell ...xP
haha it's 5 questions in 40ish minutes. So ..you don't think he would give this one? :D

Answer 40

duhhhhhhhhhhhhhhhhhh partys

Answer 41

This might be slightly faster :

Before using u-substitution, just rewrite the numerator \(x+1\) as \(\dfrac{1}{3}(3x+1)+\dfrac{2}{3}\)

Answer 42

\(\color{#000000}{ \displaystyle \int\limits_{}^{}\frac{ x+1 }{ 9x^2+6x+5 }dx }\)

\(\color{#000000}{ \displaystyle \frac{1}{3}\int\limits_{}^{}\frac{ 3x+1 }{ 9x^2+6x+5 }dx }\)

u=3x
du/3=dx

\(\color{#000000}{ \displaystyle \frac{1}{3}\int\limits_{}^{}\frac{ u+1 }{ u^2+2u+1+4 }dx }\)

\(\color{#000000}{ \displaystyle \frac{1}{6}\int\limits_{}^{}\frac{ 2u+2 }{ u^2+2u+1+4 }dx }\)

Answer 43

oh, I have some erro here.

Answer 44

\(\color{#000000}{ \displaystyle \frac{1}{6}\int\limits_{}^{}\frac{ 2u+\frac{2}{3} }{ u^2+2u+1+4 }dx }\)

i divided by 3, so it is 2/3's not 2.

Answer 45

\(\color{#000000}{ \displaystyle \frac{1}{6}\int\limits_{}^{}\frac{ 2u+2 }{ u^2+2u+1+4 }dx +\frac{1}{6}\int\limits_{}^{}\frac{ \frac{4}{3} }{ u^2+2u+1+4 }dx }\)

Answer 46

So 1 is substitution coming down to ln, and 2 is arctangent

Answer 47

k, let me see if I understand all that haha. ahh.

Answer 48

and I was wrong with that again

Answer 49

2/3=2+x
2/3-2=x
2/3-6/3=x
-4/3=x

Answer 50

\(\color{#000000}{ \displaystyle \frac{1}{6}\int\limits_{}^{}\frac{ 2u+2 }{ u^2+2u+1+4 }dx -\frac{1}{6}\int\limits_{}^{}\frac{ \frac{4}{3} }{ u^2+2u+1+4 }dx }\)

Answer 51

I overworked myself after bastekball.

Answer 52

bye:)

Answer 53

thanks for all your help:)

Answer 54

yw, whether I've done anything helpful or not:)

Answer 55

lol. Well, hopefully this isn't on the exam. =.=

Answer 56

Anyone else have ideas...?