Question

if you need to evaluate the integral but its in fraction form...
how can one do this ?

Problem:
∫ xdx / (7x^2 + 3)^5

Answer 1

Have you tried a u substitution?

Answer 2

Try, u = 7x^2+3 and see what you get.

Answer 3

no i haven't tried u substition

Answer 4

okay so u'/u right?

Answer 5

u = 7x^2+3
du=14xdx
du/14 = x dx

Answer 6

okay that looks like what i got, except how did you get du/14 =xdx ?

Answer 7

du/dx = 14x
du/14 = x dx

Answer 8

i mean why do you need to do that?

Answer 9

1/14 is just a constant, you can take this out and x dx is in your numerator you want du to equal that.

Answer 10

i see so i will plug that in to my numerator..what about the x besides the dx in the numerator ?

Answer 11

du/14 = xdx
So in your numerator xdx becomes du/14 and now you'll be integrating respect to du.

Answer 12

okay so I will integrate as I normally do ? can you show me one part to this process so i can get an idea?

Answer 13

du/14/(7x^2 + 3)^5 ?

Answer 14

\[\int\limits \frac{ x }{ (7x^2+3)^5 }dx~~~u=7x^2+3\implies du = 14xdx \implies \frac{ du }{ 14 }=xdx\]

Answer 15

\[\int\limits \frac{ 1 }{ (u)^5 } \frac{ du }{ 14 } \implies \frac{ 1 }{ 14 } \int\limits \frac{ du }{ u^5 }\]

Answer 16

Now integrate 1/u^5

Answer 17

hmmm okay!

1/u^5 is 1/u^6?

Answer 18

mhm?

Answer 19

or is it u^6/6

Answer 20

\[\int\limits \frac{ 1 }{ u^5 } du \implies \int\limits u^{-5} du \implies \frac{ u^{-5+1} }{ -5+1 }+C\]

Answer 21

Now plug u back in

Answer 22

okay okay, completely forgot about it being in the denominator means a negative exponent

Answer 23

No worries :)

Answer 24

plug it back into what now? I'm sorry i'm slow..:(

Answer 25

the equation!

Answer 26

\[\frac{ 1 }{ 14 }\frac{ u^{-4} }{ -4 }+C \implies \frac{ 1 }{ 14 }\frac{ 1 }{ -4(u)^4 } \implies \frac{ 1 }{ 14 }\frac{ 1 }{ -4(7x^2+3) }+C\] forgot to bring the 1/14 earlier :P bring that with you

Answer 27

So your final answer is \[-\frac{ 1 }{ 56(7x^2+3 )}+C\]

Answer 28

Did that make sense?

Answer 29

Oh okay thank you so mcuh @iambatman

Answer 30

Your welcome :)

Answer 31

honestly i need more practice with u-substitution but yes you made much more easier !

Answer 32

It takes a while to get used to, but no problem! :)

Answer 33

ok good i did have one last question

Answer 34

go ahead

Answer 35

ok when you plug the du back into the equation, i noticed the denominator part of it sort of "disappeared" ?

Answer 36

It didn't disappear we just made a substitution for all the stuff with x in it.