Question

Integral of (3x-10)/(x^2+8x+15)?

I got (3/2)x^2 -10x + arctan(x+4)+C but it says I'm wrong. Can anyone help me with this?

Answer 1

\[\int\limits_{}^{} (3x-10)/(x^2+8x+16-1)dx\]\[\int\limits_{}^{} (3x-10)/((x+4)^2-1)dx\]\[\int\limits_{}^{} (3x-10)dx + \int\limits_{}^{} 1/((x+4)^2-1)dx\]

This is how I did

Answer 2

your mistake is between line 2 and 3

Answer 3

\[{3x-10\over(x+4)^2-1}\neq3x-10+{1\over(x+4)^2-1}\]writing what you did clearly like this it should now seem obvious that that was not a legal move ;)

Answer 4

@Dumboy still there?

Answer 5

yeah I see my mistake, I fixed it got to this part
\[\int\limits_{}^{} 3x/((x+4)^2-1) dx -\int\limits_{}^{} 10/((x+4)^2-1) dx\]
but I'm stuck on integral..
Is this the right way to approach this?

Answer 6

sure, there are a number of ways to proceed fro here
let's do each integral individually, starting with the second one which can be rewritten\[10\int{dx\over(x+4)^2-1}\]we can use a trig sub for this I think
wanna try one?

Answer 7

Yeah that's arctan(x-4) +C but I don't get the first part of the integral. I tried trig substitution but ended up with more completed looking integral :(

Answer 8

...that integral above does not come out to arctan(x+4)+C
you would need a + in between the terms in the denom for that to work

Answer 9

\[\int{dx\over u^2-1}=-\tanh^{-1}u\]notice the \(hyperbolic\) tan there

Answer 10

+C

Answer 11

yeah these integrals are getting ugly
what section are you doing?
partial fractions maybe?

Answer 12

yeah partial frations

Answer 13

fractions*

Answer 14

ok, then let's back this up and start over, because we are set up to do a trig sub here

Answer 15

man those typos are gonna kill me...
factor instead of complete the square\[\int{3x-10\over x^2+8x-15}dx=\int{3x-10\over(x+3)(x+5)}dx\]now what do you think?
more doable yet?

Answer 16

ah I did it again!
typo, but I hope you get the point :/

Answer 17

\[\int{3x-10\over x^2+8x+15}dx=\int{3x-10\over(x+3)(x+5)}dx\]there!

Answer 18

I guess you could split it into ∫ (3x-10)/(x+3)dx + ∫(3x-10)/(x+5)dx but I'm not sure what to do next

Answer 19

no, that's not how partial fractions works at all bro :P
first of all that's not even true
second of all... brb

Answer 20

is it like this?
3x-10 = A(x-3) + B(x+5)

Answer 21

yes, much better
but the step before that is important

Answer 22

\[{3x-10\over(x+3)(x+5)}=\frac A{x+3}+\frac B{x+5}\]multiply everything by \((x+3)(x+5)\) to arrive at\[3x-10=A(x+5)+B(x+3)\]if you don't do that middle step you will be likely to lose track of which number is \(A\) and which is \(B\)

Answer 23

(btw you had x-3 as one of your factors, which is wrong.
probably my fault as I think I made that typo earlier)

Answer 24

so now that we have\[3x+10=A(x+5)+B(x+3)\]do you know how to determine A and B ?

Answer 25

when x = -3, A = -19/2
and when x = -5, B= 25/2

so (-19/2)(3x-10)/(x+3) +(25/2)(3x-10)/(x+5)?

Answer 26

idea is right, math is wrong

Answer 27

oops it's A = 2 and B = -2

Answer 28

\[3x+10=A(x+5)+B(x+3)\]\[x=-3:3(-3)+10=A(-3+5)+B(3-3)\]\[1=2A\implies A=\frac12\]

Answer 29

oh..and for B I have -5/2 but shouldn't I get -1/2?

Answer 30

yeah I think you should get -1/2
let's see...

Answer 31

\[x=-5:3(-5)+10=A(-5+5)+B(-5+3)\]\[-5=-2B\implies B=\frac52\]oh, so we were both wrong :P

Answer 32

\[1/2 \int\limits_{}^{} (3x-10)/(x+3)dx\]\[3x-10 \div x+3 = 3-1/(x+3)\]\[\int\limits\limits_{}^{} 3 - 1/(x+3) dx\]\[3x+\ln(x+3)+C\]

is this part correct?

Answer 33

no, sorry

Answer 34

I forgot 1/2 in the last part, is that what's wrong or is it other part?

Answer 35

I'm sorry but you seem to have no idea what we're doing here
you didn't use all the stuff we just did with partial fractions
where is the x+5 term?
why do you still have 3x-10 ?
that should went away when we did the PF part

Answer 36

I will demonstrate this one and link you to something I hope that you read thoroughly to clear up these misunderstandings...

Answer 37

\[\int{3x-10\over x^2+8x-15}dx=\int{3x-10\over(x+3)(x+5)}dx=\int\frac A{x+3}+\frac B{x+5}dx\]partial fractions:\[A(x+5)+B(x+3)=3x-10\]\[x=-3\implies A=-\frac12\]\[x=-5\implies B=\frac52\]so the integral is now\[-\frac12\int\frac 1{x+3}dx+\frac52\int\frac 1{x+5}dx\]do you understand now?

Answer 38

oh okay so 3x-10 goes away with partial fraction..
I was just doing the first part of the integral up there (forgot minus sign again:P)
Thanks