Integration by parts help

There's an integral I don't know how to do and I'm told that I'm to use integration by parts. It's the integral of x^7 divided by sqrt(x^4 + 9).
I know that if I let u = x^2 / 3 then I have 27/2 multiplied by the integral of u^3 divided by sqrt(u^2 + 1). But I don't know where to go from here.

Scroll to see replies

Reply 1

Zacken

Original post by UpstairsMuffin

Try using integration by parts with

f = u^2

and

\displaystyle \mathrm{d}v = \frac{u}{\sqrt{u^2 + 1}}

(this is a standard form f'(x)f(x)^n to integrate.

Reply 2

username1426065

Original post by Zacken

Try using integration by parts with

f = u^2

and

\displaystyle \mathrm{d}v = \frac{u}{\sqrt{u^2 + 1}}

(this is a standard form f'(x)f(x)^n to integrate.

Then I end up with the integral of (u^2 + 1)^3/2. :/

(edited 8 years ago)

Reply 3

Zacken

Original post by UpstairsMuffin

Then I end up with the integral of (u^2 + 1)^3/2. :/

Did you? Mind showing me your working? That doesn't seem correct.

Reply 4

username1426065

Original post by Zacken

Did you? Mind showing me your working? That doesn't seem correct.

With a whole bunch of other stuff added to it from the IBP, I mean.

Reply 5

Zacken

Original post by UpstairsMuffin

With a whole bunch of other stuff added to it from the IBP, I mean.

It's the whole other stuff that makes it very easy to integrate.

Reply 6

TeeEm

Original post by Zacken

It's the whole other stuff that makes it very easy to integrate.

go and do some STEP papers ... otherwise I can see you in Warwick ...
(you spend too long in here)

Reply 7

username1426065

Original post by Zacken

It's the whole other stuff that makes it very easy to integrate.

The whole other stuff is outside of the integral.

Reply 8

username1426065

Original post by TeeEm

go and do some STEP papers ... otherwise I can see you in Warwick ...
(you spend too long in here)

I've done lots of integrals, I just find this one very difficult. :frown:

Reply 9

Zacken

Original post by TeeEm

go and do some STEP papers ... otherwise I can see you in Warwick ...
(you spend too long in here)

I'm not at home, having dinner out! Can't do STEP right now. :tongue:

(I do though, you have a point)

Original post by UpstairsMuffin

The whole other stuff is outside of the integral.

Please show us your working, the other stuff should not be outside the integral. You should have

\int \frac{d}{du} (u^2) \cdot (\text{your thing to the power 3/2}) \, \mathrm{d}u

inside the integral. Sorry for the bad LaTeX, it's very hard to type on a phone.

Reply 10

tinkerbella~

Original post by Zacken

It's the whole other stuff that makes it very easy to integrate.

Talk to your family I'm sure they're nice people.

Reply 11

username1426065

Original post by Zacken

I'm not at home, having dinner out! Can't do STEP right now. :tongue:

(I do though, you have a point)

Please show us your working, the other stuff should not be outside the integral. You should have

\int \frac{d}{du} (u^2) \cdot (\text{your thing to the power 3/2}) \, \mathrm{d}u

inside the integral. Sorry for the bad LaTeX, it's very hard to type on a phone.

Reply 12

Zacken

Original post by UpstairsMuffin

Huh? You're using integration by parts on with

f = u^2 \Rightarrow \frac{df}{du} = 2u

and

\mathrm{d}v = \int \frac{u}{\sqrt{u^2 + 1}}

So using the integration by parts formula (look this up, I think you're getting it wrong!) you should get:

\displaystyle [u^2 \cdot (u^2 + 1)^{3/2} - \int 2u (u^2 + 1)^{3/2} \, \mathrm{d}u

(I really hope that latex works...)

Reply 13

username1426065

Original post by Zacken

Huh? You're using integration by parts on with

f = u^2 \Rightarrow \frac{df}{du} = 2u

and

\mathrm{d}v = \int \frac{u}{\sqrt{u^2 + 1}}

So using the integration by parts formula (look this up, I think you're getting it wrong!) you should get:

\displaystyle [u^2 \cdot (u^2 + 1)^{3/2} - \int 2u (u^2 + 1)^{3/2} \, \mathrm{d}u

(I really hope that latex works...)

Aha! Thank you! I just got confused over what I was differentiating and what I was antidifferentiating. It's easy to lose track of what you're doing when you're trying to do so many things in your head at once. That one mistake caused me a lot of pain. That's it now, thanks.

Reply 14

Zacken

Original post by UpstairsMuffin

De nada. Glad I helped.

Reply 15

Indeterminate

Original post by TeeEm

go and do some STEP papers ... otherwise I can see you in Warwick ...
(you spend too long in here)

Even I was keen not to end up at Warwick; the university just seems a bit low-profile in comparison to Oxbridge, Imperial, UCL and so on.

But our @Zacken will definitely end up at Cambridge :yep:

Reply 16

Student403

Original post by Zacken

I'm not at home, having dinner out! Can't do STEP right now. :tongue:

(I do though, you have a point)

Please show us your working, the other stuff should not be outside the integral. You should have

\int \frac{d}{du} (u^2) \cdot (\text{your thing to the power 3/2}) \, \mathrm{d}u

inside the integral. Sorry for the bad LaTeX, it's very hard to type on a phone.

Omg you're having dinner out and you're on your phone helping out on the maths forum?

Definition of dedication xD (or addiction :erm:

)

Reply 17

TeeEm

Original post by Indeterminate

Even I was keen not to end up at Warwick; the university just seems a bit low-profile in comparison to Oxbridge, Imperial, UCL and so on.

But our @Zacken will definitely end up at Cambridge :yep: