[SOLVED] Evaluating this indefinite integral

**fattydq** · Mar 3rd 2009, 11:53 AM

The indefinite integral of x^3sq root of x^2+1. I have no idea where to even start. I'm guessing I'd have to use substitution of some sort to find this antiderivative but I don't know how the square root would come into play. I know I could use x^3 as u and du would then be 3x^2 so it'd be 1/3 the integral of the square root of 1 which is just as complicated.

**galactus** · Mar 3rd 2009, 12:55 PM

You could use trig sub.

$\int x^{3}\sqrt{x^{2}+1}dx$

Let $x=tan(t), \;\ dx=sec^{2}(t)dt$

Upon making the subs, we get:

$\int tan^{3}(t)sec^{3}(t)dt$

$\int tan^{2}(t)sec^{2}(t)(sec(t)tan(t)dt$

= $\int (sec^{2}(t)-1)sec^{2}(t)(sec(t)tan(t))dt$

Let $u=sec(t), \;\ du=sec(t)tan(t)dt$

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

$\frac{1}{5}u^{5}-\frac{1}{3}u^{3}$

Resub:

$\frac{1}{5}sec^{5}(t)-\frac{1}{3}sec^{3}(t)$

Resub from the beginning, $t=tan^{-1}(x)$

This gives:

$\frac{1}{5}(x^{2}+1)^{\frac{5}{2}}-\frac{1}{3}(x^{2}+1)^{\frac{3}{2}}+C$

~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~ ~~~~~~~~~~~~~~~~~~~~~~

An easier way may be to let $u=x^{2}+1, \;\ du=2xdx, \;\ \frac{1}{2}du=dx$

Then we get:

$\frac{1}{2}\int \sqrt{u}(u-1)du=\frac{1}{2}\left[\int u^{\frac{3}{2}}-\int u^{\frac{1}{2}}\right]du$

Can you go from there?. This way is easier if the trig sub is scary.

**Reckoner** · Mar 3rd 2009, 12:59 PM

You can also use integration by parts, with $u=x^2$ and $dv=2x\sqrt{x^2+1}\,dx$ :

$\int x^3\sqrt{x^2+1}\,dx$

$=\frac12\int x^2\left(2x\sqrt{x^2+1}\right)dx$

$=\frac12\int x^2\left[\frac23(x^2+1)^{3/2}\right]'dx$

$=\frac12\left[\frac23x^2(x^2+1)^{3/2}-\frac23\int2x(x^2+1)^{3/2}\,dx\right]$

$=\frac13x^2(x^2+1)^{3/2}-\frac13\int2x(x^2+1)^{3/2}\,dx$

$=\frac13x^2(x^2+1)^{3/2}-\frac13\left[\frac25(x^2+1)^{5/2}\right]+C$

$=\frac13x^2(x^2+1)^{3/2}-\frac2{15}(x^2+1)^{5/2}+C$

**Krizalid** · Mar 3rd 2009, 01:03 PM

Just put $u^2=x^2+1$ !

**fattydq** · Mar 3rd 2009, 02:24 PM

Originally Posted by galactus

You could use trig sub.

$\int x^{3}\sqrt{x^{2}+1}dx$

Let $x=tan(t), \;\ dx=sec^{2}(t)dt$

Upon making the subs, we get:

$\int tan^{3}(t)sec^{3}(t)dt$

$\int tan^{2}(t)sec^{2}(t)(sec(t)tan(t)dt$

= $\int (sec^{2}(t)-1)sec^{2}(t)(sec(t)tan(t))dt$

Let $u=sec(t), \;\ du=sec(t)tan(t)dt$

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

$\frac{1}{5}u^{5}-\frac{1}{3}u^{3}$

Resub:

$\frac{1}{5}sec^{5}(t)-\frac{1}{3}sec^{3}(t)$

Resub from the beginning, $t=tan^{-1}(x)$

This gives:

$\frac{1}{5}(x^{2}+1)^{\frac{5}{2}}-\frac{1}{3}(x^{2}+1)^{\frac{3}{2}}+C$

~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~ ~~~~~~~~~~~~~~~~~~~~~~

An easier way may be to let $u=x^{2}+1, \;\ du=2xdx, \;\ \frac{1}{2}du=dx$

Then we get:

$\frac{1}{2}\int \sqrt{u}(u-1)du=\frac{1}{2}\left[\int u^{\frac{3}{2}}-\int u^{\frac{1}{2}}\right]du$

Can you go from there?. This way is easier if the trig sub is scary.

I'm sorry, I don't understand where any of those numbers came from. Because there is no x in the function, they are all x's raised to powers, and none of your u's are raised to powers?

I haven't learned any of the methods any one else used yet, other than substitution to solve this problem. So if you guys, when helping me, could focus on the substitution method that would be appreciated.

**fattydq** · Mar 3rd 2009, 02:25 PM

Originally Posted by Krizalid

Just put $u^2=x^2+1$ !

but then what would du be? And then what?

**fattydq** · Mar 3rd 2009, 03:11 PM

Originally Posted by Reckoner

You can also use integration by parts, with $u=x^2$ and $dv=2x\sqrt{x^2+1}\,dx$ :

$\int x^3\sqrt{x^2+1}\,dx$

$=\frac12\int x^2\left(2x\sqrt{x^2+1}\right)dx$

$=\frac12\int x^2\left[\frac23(x^2+1)^{3/2}\right]'dx$

$=\frac12\left[\frac23x^2(x^2+1)^{3/2}-\frac23\int2x(x^2+1)^{3/2}\,dx\right]$

$=\frac13x^2(x^2+1)^{3/2}-\frac13\int2x(x^2+1)^{3/2}\,dx$

$=\frac13x^2(x^2+1)^{3/2}-\frac13\left[\frac25(x^2+1)^{5/2}\right]+C$

$=\frac13x^2(x^2+1)^{3/2}-\frac2{15}(x^2+1)^{5/2}+C$

I'm sorry, I don't want to solve it that way as I haven't learned that method yet

**skeeter** · Mar 3rd 2009, 03:38 PM

Originally Posted by fattydq

I'm sorry, I don't want to solve it that way as I haven't learned that method yet

$\int x^3 \sqrt{x^2 + 1} \, dx$

$u = x^2 + 1<br />$

$x^2 = u - 1$

$du = 2x \, dx$

set up the original integral for substitution ...

$\frac{1}{2} \int x^2 \sqrt{x^2 + 1} \cdot 2x \, dx$

substitute ...

$\frac{1}{2} \int (u - 1)\sqrt{u} \, du$

can you finish?

**fattydq** · Mar 3rd 2009, 03:39 PM

I solved it. Thanks guys!

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