Question

cant figure out this integral for the life of me..

Answer 1

oh good, I was getting bored

Answer 2

spent about 2 hours on this last night not sure if i did it right..\[\int\limits_{0}^{1}(2x^3+5x)/(x^4+5x^2+4) dx\]

Answer 3

ha good then. sorry if it looks funny i can never do it right... first thing i did was split up the bottom into x^2+4 and x^2+1 right?

Answer 4

\[u=x^4+5x^2+4\]

Answer 5

du=...?

Answer 6

The integral looks suspiciously like \(\displaystyle{\int\frac{f'\left(x\right)}{f\left(x\right)}}\,dx\), which yells U-sub, as Turing is pointing out.

Answer 7

\[du=4x^3+ 10x\]

Answer 8

are you sure i shouldnt factor?

Answer 9

and what can we do to that expression for it to be the same as the numerator?

Answer 10

\[du=4x^3+10xdx\implies 2x^3+5xdx=\frac{du}2\]

Answer 11

writing the dx is important (for me at least) to help me see how the u-sub will change things

Answer 12

1/2*log(x^4+5x^2+4)
putting values you would get
1/2*(log(10)-log(4))

Answer 13

whoa how did you get to that experiementx?

Answer 14

by doing what I'm talking about!!!

Answer 15

its been a while since I did one of these.. struggling

Answer 16

not to be rude, but if you experimentX is saying what wolfram alpha could have.
-the point is \(how\) to solve the problem

Answer 17

thats what confused me. I dont need the answer, i want to know how to do it.

Answer 18

When I do u sub, what does the equation become? would it be \[\int\limits_{0}^{1} du/2u?\]

Answer 19

yes

Answer 20

not the bounds though, those change

Answer 21

but the indefinite integral is right

Answer 22

So i pull out the 1/2? and then its lnu when u=x^4+5x^2+4?

Answer 23

+C
yes

Answer 24

integ f'(x)/f(x) dx = log(f(x)) + c thats the formula

Answer 25

change the limits when you make the u - sub then you don't have to switch back

Answer 26

sat, in this case since it's just one and zero it's easier to leave them in x?

Answer 27

I meant to say "dont you think"?

Answer 28

\[u(0)=4,u(1)=10\] then
\[\frac{1}{2}\int_4^{10}\frac{du}{u}\]

Answer 29

ah I got it now. I never really change the bounds just subed back in.and got 1/2(ln10-ln4)

Answer 30

of course it makes no difference, but why go back? you can just about finish this in your head
\[\frac{1}{2}(\ln(10)-\ln(4))\]

Answer 31

or if you don't change the bounds and change it back at the end\[\frac12\int\frac{du}u=\frac12\ln u=\frac12\ln(x^4+5x^2+4)|_{0}^{1}\]should give the same answer as satellite

Answer 32

yes of course same answer
i guess it is a matter of preference

Answer 33

Yeah turing thats how I did it and got the same answer obviously

Answer 34

hooray, math is still consistent :)

Answer 35

Thank you..glad i almost got this deathly homework done

Answer 36

You will most likely hear from me seen again ha