# integration problem

• Feb 6th 2008, 10:13 PM
Andreamet
integration problem
x^3*(x^2+1)^(1/2)
• Feb 7th 2008, 04:04 AM
galactus
$\int{x^{3}\sqrt{x^{2}+1}}dx$

If you let $u=x^{2}+1, \;\ \frac{du}{2}=xdx \;\ u-1=x^{2}$

it'll turn it to something much easier.
• Feb 7th 2008, 06:44 AM
Krizalid
To get rid of the square root, one also can set $u^2=x^2+1.$
• Feb 7th 2008, 06:47 AM
Soroban
Hello, Andreamet!

Quote:

$\int x^3(x^2+1)^{\frac{1}{2}}\,dx$
We can do it By Parts . . .

. . . $\begin{array}{ccccccc}u & = & x^2 & & dv & = & x(x^2+1)^{\frac{1}{2}}\,dx \\ du &=& 2x\,dx & & v &=&\frac{1}{3}(x^2+1)^{\frac{3}{2}} \end{array}$

And we have: . $\frac{1}{3}x^2(x^2+1)^{\frac{3}{2}} - \frac{2}{3}\int x(x^2+1)^{\frac{3}{2}}\,dx\quad \hdots\quad etc.$

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

Another approach is Trig Substitution . . .

Let $x \:=\:\tan\theta\quad\Rightarrow\quad dx \:=\:\sec\theta\tan\theta\,d\theta$
. . and: . $\sqrt{x^2+1} \:=\:\sec\theta$

Substitute: . $\int \tan^3\!\theta \cdot \sec\theta\,(\sec^2\!\theta\,d\theta) \;=\;\int\sec^3\!\theta\tan^3\!\theta\,d\theta$

. . $= \;\int\sec^2\!\theta\tan^2\!\theta\,(\sec\theta\ta n\theta\,d\theta) \;=\;\int\sec^2\!\theta(\sec^2\!\theta - 1)\,(\sec\theta\tan\theta\,d\theta)$

. . $= \;\int(\sec^4\!\theta - \sec^2\!\theta)\,(\sec\theta\tan\theta\,d\theta)$

Then let $u \,=\,\sec\theta$ . . .