Integrate by parts and substitution

[hank]

Ok, here's the problem:

S x^3 / sqrt(x^2 + 1) limits = 0 to 1.

Ok, here's as far as I get by integration by parts:

u = x^3
du = 3x^2
dv = 1 / sqrt(x^2 + 1)
v = ln (x + sqrt(x^2 + 1)


= x^3 ln (x + sqrt(x^2 + 1) - 3 S x^2 ln (x + sqrt(x^2 + 1) dx

Not sure where to go from here. Is my setup right?


The problem further tells me to integrate by substitution where u = sqrt (x^2 + 1)

I don't think I can do it tho because I can't do anything with the x^3.

u = sqrt(x^2 + 1)
du = x / sqrt(x^2 + 1)

Stuck here as there's nothing I can do with x^3.

Any help would be great.

 

[stapel]

Try u = sqrt[x² + 1], so du = (x/sqrt[x² + 1]) dx and u² - 1 = x². :wink:

Eliz.

 

[galactus]

This actually whittles down to an easy integral.

You don't need parts. You're making it too difficult.

As suggested, let \(\displaystyle u=\sqrt{x^{2}+1}, \;\ x^{2}=u^{2}-1, \;\ du=\frac{x}{\sqrt{x^{2}+1}}dx\)

See?. an x^2 is taken care of with the u^2-1 and du takes care of the rest.

Don't forget to change the limits of integration.

So, you have:

\(\displaystyle \L\\\int(u^{2}-1)du\)

Now integrate and resub.

 

[hank]

This actually whittles down to an easy integral.

You don't need parts. You're making it too difficult.

As suggested, let \(\displaystyle u=\sqrt{x^{2}+1}, \;\ x^{2}=u^{2}-1, \;\ du=\frac{x}{\sqrt{x^{2}+1}}dx\)

See?. an x^2 is taken care of with the u^2-1 and du takes care of the rest.

Don't forget to change the limits of integration.

So, you have:

\(\displaystyle \L\\\int(u^{2}-1)du\)

Now integrate and resub.

Hi, galactus, nice to see you again.
Unfortunately, the problem requires me to do it by both parts and u-sub.

 

[galactus]

This is more of a substitution problem. Parts isn't a very good method.

Trig sub would work good, too.

Let \(\displaystyle x=tan(u), \;\ dx=sec^{2}(u)du\)

Then you get \(\displaystyle \L\\\int_{0}^{\frac{\pi}{4}}tan(u)sec^{2}(u)du\)


BUT, here's parts:

Let \(\displaystyle \L\\u=x^{2}, \;\ dv=\frac{x}{\sqrt{x^{2}+1}}dx, \;\ du=2xdx, \;\ v=\sqrt{x^{2}+1}\)

Now assemble it with the parts.