Integrating inverse trig functions

[Jason03]

I was working on this problem

<br /> <br /> \int{\frac{x^3}{x^{2}+1}}dx<br /> <br />

I at first tried to use one of the inverse trig functions but couldn't get the form to match...should I try to use log properties...making the denominaor u and trying to get the numerator 1?

 

[Focus]

You are doing this far too complicated. Try a simple substitution. Hint: x^2+1.

 

[Jason03]

I was thinking of making u = x^2 + 1...but doesn't the numerator need to be 1?...the numerator is what's throwing me off...

 

[snipez90]

I'm not sure why the numerator needs to be 1...

As long as you can express everything in terms of u, everything should work out ok. So Focus's hint should solve the problem.

 

[Jason03]

if I do the substitution...i get 1/2du = xdx...but the problem has x^3 dx...?

 

[Dick]

x^3*dx=x^2*(x*dx). Now express x^2 in terms of u.

 

[Jason03]

I think I see what your saying...so

1/2du = xdx so x^2 = u - 1

so would this integral be true?

<br /> <br /> \int{\frac{u-1}{u}*\frac{du}{2}}<br /> <br />

 

[Dick]

That's the one.

 

[th3plan]

Hey just my two thoughts , just expand the integral and u get then x-x/(x^2 +1 )

then its very simple, but u can do it your way also, same answer u will arrive to
. They taught me this in Calculus II, this method, don't remember what its called though heh.

 

[Dick]

Hey just my two thoughts , just expand the integral and u get then x-x/(x^2 +1 )

then its very simple, but u can do it your way also, same answer u will arrive to
. They taught me this in Calculus II, this method, don't remember what its called though heh.

It's called polynomial division. It's about the same amount of work either way (I think).

 

[Jason03]

ok...I got close to the correct answer but still not there...


I took the above integral with the u functions and separated them

integral of 1/2(u/u) - 1/2(1/u)



the answer is (1/2)x^2 - (1/2)ln(x^2+1)


I understand the integration of the second term with the natural log...but not the first term...

 

[Focus]

ok...I got close to the correct answer but still not there...


I took the above integral with the u functions and separated them

integral of 1/2(u/u) - 1/2(1/u)



the answer is (1/2)x^2 - (1/2)ln(x^2+1)


I understand the integration of the second term with the natural log...but not the first term...

The first term is just the integral of 1/2 with respect to u, so it is 1/2u=1/2(x^2+1)

 

[Jason03]

well than how does that get you the solution of 1/2(x^2) for the first term??

 

[Focus]

well than how does that get you the solution of 1/2(x^2) for the first term??

Don't forget the constant when you integrate so the real answer is 1/2(x^2)+1/2l+1/2og(x^2+1)+k (let c=k+1/2) and voila