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integrating help

I am having trouble integrating this

(x^3)/(1+x^2)^(1/2)

i know that i let u = 1+x^2
u'=2x

i have no idea what to do now some help would be great thanks.
Original post by luke123456
I am having trouble integrating this

(x^3)/(1+x^2)^(1/2)

i know that i let u = 1+x^2
u'=2x

i have no idea what to do now some help would be great thanks.


Try this substitution instead, it will be easier: u=x^2
Original post by jameswhughes
Try this substitution instead, it will be easier: u=x^2


Are you sure about that? the one the OP posted looks easier.
Original post by ghostwalker

Original post by ghostwalker
Are you sure about that? the one the OP posted looks easier.


Yes, I've just done it-OP's substitution gives something which is hard to integrate.

Edit: oops, my mistake, they both work equally well :smile:
(edited 12 years ago)
Original post by luke123456
I am having trouble integrating this

(x^3)/(1+x^2)^(1/2)

i know that i let u = 1+x^2
u'=2x

i have no idea what to do now some help would be great thanks.


Sorry if this is pre-empting, but thread seems to have stalled.

Since this is the one you've started with, you just need to substitute in.

Notice that x3=x2×xx^3 = x^2 \times x

See what you can do.
Reply 5
Avatar for ztibor
ztibor
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Original post by luke123456
I am having trouble integrating this

(x^3)/(1+x^2)^(1/2)

i know that i let u = 1+x^2
u'=2x

i have no idea what to do now some help would be great thanks.


Do not substitute. It will be more simple with some modification
x31+x2=1+x2−11+x2⋅x=12⋅2x1+x2−12⋅2x1+x2\frac{x^3}{\sqrt{1+x^2}}=\frac{1+x^2-1}{\sqrt{1+x^2}}\cdot x=\frac{1}{2}\cdot 2x\sqrt{1+x^2}-\frac{1}{2}\cdot \frac{2x}{\sqrt{1+x^2}}

both terms can be integrated with rule of ∫f′(x)⋅[f(x)]ndx=fn+1(x)n+1+C\int f'(x)\cdot [f(x)]^n dx=\frac{f^{n+1}(x)}{n+1}+C

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