Question

Need help on integration

Answer 1

\[\frac{ x }{ \sqrt{5+4x}}dx\]

Answer 2

\[\Large\rm u=5+4x\qquad\to\qquad du=4dx\qquad\to\qquad \frac{1}{4}du=dx\]
Then you also need to do a little work with this first substitution part that we defined,\[\Large\rm u=5+4x\qquad\to\qquad u-5=4x\qquad\to\qquad \frac{1}{4}(u-5)=x\]

Answer 3

\[\Large\rm \int\limits \frac{x}{\sqrt{5+4x}}dx\quad=\quad \]Understand how we can use these three pieces to make our substitution?\[\Large\rm 5+4x=u\]\[\Large\rm x=\frac{1}{4}(u-5)\]\[\Large\rm dx=\frac{1}{4}du\]

Answer 4

Is this to be solved by Integration of parts?

Answer 5

You certainly can do parts `instead` if you prefer that method :)

Answer 6

This method avoids doing parts though.

Answer 7

That;s what we are covering, thanks.

Answer 8

But I follow on what you were showing above.

Answer 9

So if you wanted to do parts, you would do something like...\[\Large\rm \int\limits\limits \frac{x}{\sqrt{5+4x}}dx\quad=\quad\int\limits x\left(\frac{1}{\sqrt{5+4x}}dx\right)\]\[\Large\rm u=x,\qquad\qquad dv=\frac{1}{\sqrt{5+4x}}dx\]Finding your v is a little tricky

Answer 10

I would make the substitution a=5+4x for your dv

Answer 11

And integrate it like so with the power law

Answer 12

OK I did break the x out but didn't use the correct dv.

Answer 13

Oh did you use \(\Large\rm dv=\sqrt{5+4x}\) by mistake maybe? :3

Answer 14

2(5+4x)^1/2 ?

Answer 15

What is that? Your v?

Answer 16

Yes

Answer 17

Mmm close, I think maybe you forgot about the 4 on the x?
You should be getting a 1/4 showing up somewhere.
Then your stuff to the -1/2 power becomes 1/2 power...
and divide by 1/2, multiply by 2.

I think you need an 8 in front, not a 2.

Answer 18

You can always do a substitution on the side (as Kieth suggested), to get your dv figured out.

Answer 19

From the 4x correct

Answer 20

8(5+4x)^1/2

Answer 21

Woops, not 8 sorry sorry.
Multiplying by 1/4 (from the 4x).
So your 2 with the 1/4 gives us ... 1/2 in front.
Ahh my bad >.<

Answer 22

\[\Large\rm v=\frac{1}{2}(5+4x)^{1/2}\]I can go into clearer detail if you need.
I'm just not sure where you are at with your integrating skills right now,
so maybe I'm assuming too much.

Answer 23

\[1/2(x)(5+4x)^{\frac{ 1 }{ 2 }}-\frac{ 1 }{ 12 }(5+4x)^{\frac{ 3 }{ 2 }}+C\]

Answer 24

You integrated your v du already?
Mmm great! Looks correct! :)

Answer 25

Thanks you are right I did try dv=sq root (5+4x)

Answer 26

This problem is simpler with the substitution:

\(\large 5 + 4x = u^2; ~~~~ 4dx = 2udu; ~~~~ x = (u^2-5)/4\)

Answer 27

I agree but we are covering integration by parts.

Answer 28

Sure. This was just a side note.