Question

Integral Help!

Answer 1

Find the integral for...\[\int\limits_{0}^{\infty} (x/ (x^2 + 1)(3x +1))dx\]

Answer 2

ick

Answer 3

looks like you need partial fractions right?

Answer 4

*sigh*

Answer 5

whats ick mean

Answer 6

ick means we have to use partial fractions to solve this integral

Answer 7

Or actually I wrote the integral wrong..

Answer 8

o-o

Answer 9

i am not above cheating to do this donkey work

Answer 10

http://www.wolframalpha.com/input/?i=x%2F%28%28x^2%2B1%29%283x%2B1%29%29

Answer 11

\[\int\limits_{0}^{\infty} ((x-3)/ (x^2+1) (3x+1))dx\]

Answer 12

did you just change the problem?

Answer 13

Yah, sorry! I wrote it wrong! :(

Answer 14

I'm 100% it's right this time lolxD

Answer 15

I just need to show that this converges. Which means to basically evaluate it to show it goes to a finite number.

Answer 16

much much (much) bettter

Answer 17

Hahaha :) Shall we use limits? :D

Answer 18

\[\int \frac{x}{x^2+1}dx-\int \frac{3}{3x+1}dx\] get two logs

Answer 19

the denominator would bee x^2+1 * 3x+1

Answer 20

mental u - sub gives
\[\frac{1}{2}\log(x^2+1)-\log(3x+1)\]

Answer 21

ok did i go too fast, we can back up

Answer 22

first partial fractions

Answer 23

Yes please! :) ok

Answer 24

ok lets do it right

Answer 25

Hahaha (:

Answer 26

\[\frac{x+3}{(x^2+1)(3x+1)}=\frac{x}{x^2+1}-\frac{3}{3x+1}\]

Answer 27

that via partial fraction decomposition

Answer 28

okay this is the part I'm confused on. How did you receive that?

Answer 29

that is why i said "ick"

Answer 30

Ahhh I see.

Answer 31

\[\frac{Ax+B}{x^2+1}+\frac{C}{3x+1}\] is a start

Answer 32

multiply it all by (x^2 +1) (3x+1)!

Answer 33

then
\[(Ax+B)(3x+1)+C(x^2+1)=x+3\] solve for A, B, C

Answer 34

omg what grade is this homewwork omggggggggg

Answer 35

Calculus 1.5 ish

Answer 36

im only in 6th grade holy crap

Answer 37

It would be x-3! :)

Answer 38

yes

Answer 39

so
\[\int \frac{x}{x^2+1}dx-\int \frac{3}{3x+1}dx\]

Answer 40

both are very easy, mental u - sub

Answer 41

\[\frac{1}{2}\log(x^2+1)-\log(3x+1)\]

Answer 42

now before we try to take the limit as \(x\to \infty\) write as a single log

Answer 43

Wait can you back up on the u substitution please? :)

Answer 44

ok

Answer 45

first one, \(u=x^2+1,du=2xdx,\frac{1}{2}du=xdx\) gives
\[\frac{1}{2}\int \frac{du}{u}\] this should be a mental step

Answer 46

that gives
\[\frac{1}{2}\log(u)=\frac{1}{2}\log(x^2+1)\]

Answer 47

Okay I understand now

Answer 48

second one also a mental u - sub \(u=3x+1, du=3dx\) \[\int\frac{du}{u}=\log(u)=\log(3x+1)\]

Answer 49

now we are here \[\frac{1}{2}\log(x^2+1)-\log(3x+1)\] right?

Answer 50

yes! That makes so much sense

Answer 51

but before we take the limit as x goes to infinity, we have to write as a single log

Answer 52

otherwise it will look like \(\infty-\infty\)

Answer 53

I see

Answer 54

via properites of the log you get
\[\log\left(\frac{\sqrt{x^2+1}}{3x+1}\right)\]

Answer 55

OMG THAT IS GENIUS.

Answer 56

why thank you!

Answer 57

take
\[\lim_{x\to \infty}\log\left(\frac{\sqrt{x^2+1}}{3x+1}\right)\] get \[\log(\frac{1}{3})=-\log(3)\] in your head

Answer 58

That was quite an adventure! Thank you

Answer 59

on account of
\[\lim_{x\to \infty}\left(\frac{\sqrt{x^2+1}}{3x+1}\right)=\frac{1}{3}\]with no agony

Answer 60

hope you enjoyed the ride
good luck
yw