Question

Calculate the indefinite integral: Integral (2x)/(x+3)(3x+1)

Answer 1

well how far have you gotten?

Answer 2

i have no idea how to do this one

Answer 3

it's the exact same thing

Answer 4

actually it's even easier because the denominator is already factored

Answer 5

how do you split up the denominator in partial fractions (PF) ?

Answer 6

o i see so a/x+3 + b/3x+1 =2x

Answer 7

actually it's
a/x+3 + b/3x+1 =2x/(x+3)(3x+1)
an easy mistake to make, but don't
you are confusing what you wrote with the result of the next step, which is to multiply that by (x+3)(3x+1) on both sides

Answer 8

see, the first step is to split the fraction up\[\frac{2x}{(x+3)(3x+1)}=\frac A{x+3}+\frac B{3x+1}\]

Answer 9

so a(3x+1)+b(x+3)=2x/(x+3)(3x+1)?

Answer 10

no, now the part on the right cancels because we multiply *both sides* by (x+3)(3x+1), no *now* it's\[A(3x+1)+B(x+3)=2x\]gotta make sure you understand each step. The mistakes you are making may seem small but can mess you up really bad if you're not careful

Answer 11

kk so i finished subsitution and got a =2/3 and b=-1/4

Answer 12

\[\frac{2x}{(x+3)(3x+1)}=\frac A{x+3}+\frac B{3x+1}\]\[2x=A(3x+1)+B(x+3)\]ok let me check...

Answer 13

I think I'm getting different numbers

Answer 14

i used -3 and -1/3

Answer 15

just doing -3 I am getting something else

Answer 16

-3 makes b cancel out and makes -8a=-6?

Answer 17

and that is not =2/3

Answer 18

3/4 -.-

Answer 19

haha, it happens to the best of us :)

Answer 20

haha got it now. but for b im certain i get -1/4

Answer 21

o wait no never mind

Answer 22

yeah that's what I get
I think you can handle it from here

Answer 23

nevermind i got ithaha

Answer 24

I think you've gotten the hang of basic PF problems, congrads :)

Answer 25

thanks a lot!, can you give me two minutes, i would just like to check my answer with you

Answer 26

sure

Answer 27

i got 3ln+9/4ln-ln

Answer 28

I think you may have switched A and B in your integrand

Answer 29

also I think you did the u-sub wrong

Answer 30

isnt it 3/4ln(x+3) - 1/4ln(3x+1) 1-0
there abs value i know

Answer 31

second term needs a u-sub ;)

Answer 32

okay now im lost haha

Answer 33

\[-\frac14\int\frac1{3x+1}dx\]\[u=3x+1\implies du=3dx\implies dx=\frac13du\]\[-\frac1{12}\int\frac1udu=-\frac1{12}\ln u+C\]

Answer 34

though must not forget calculus 1 in calculus 2

Answer 35

i still dont understand why we have to do that though

Answer 36

you have become rusty on your u-substitutions, we need to sub the integrand completely in terms of u and du, so we have to make some algebraic manipulation to get an expression for dx

Answer 37

but how would i know when to do that?

Answer 38

forget the -1/4 for a second\[\int\frac1{3x+1}dx\]\[u=3x+1\implies du=3dx\]here is how we know when to do that: notice we don't have a 3dx in our integrand, we just have dx, so we know we need to solve our expression above for dx\[dx=\frac13du\]no we can sub in for each part of the integrand completelu in terms of u and du\[\int\frac1{3x+1}dx=\int\frac1u(\frac13du)=\frac13\int\frac1udu\]

Answer 39

yes i understand that but then how come you didnt do the same for the first part with the 3/4 ln(x+3)

Answer 40

you can if you want but it is a waste of time because the coefficient of x is 1\[\int\frac1{x+3}dx\]\[u=x+3\implies du=dx\]\[\int\frac1udu\]completely pointless, we already had a dx in our integrand we needed no algebraic manipulation to get it

Answer 41

ooooooo!!!! so pretty much when to coefficient is greater than 1 you have to do the u sub rule!

Answer 42

well almost, but not exactly. You should always check
for instance, what id instead of A=-1/4 we had A=3
we would then have...

Answer 43

you mean b=-1/4?

Answer 44

right, B, if it were 3 and not -1/4\[3\int\frac1{3x+1}dx\]\[u=3x+1\implies du=3dx\]now we can just bring the 3 into the integral and notice that\[\int\frac{3dx}{3x+1}=\int\frac{du}u=\ln u+C=\ln|3x+1|+C\]you still have to make sure your substitution is complete, however you want to look at it.

Answer 45

yeh i understand :) and now so far we have 3/4ln(x+3)-1/12ln(3x+1) + C

Answer 46

and we're done

Answer 47

now subsitution of 0 and 1?

Answer 48

I thought you said it was indefinite

Answer 49

yes. yes it is haha.. i'm really off today haha

Answer 50

it's all good :)
I gotta go, see ya!

Answer 51

nono can you help me with one more problem

Answer 52

I can't, really. Next time friend!