Integral x^4(4+x^5)^1/2

**cammywhite** · April 18th 2009, 09:41 PM

$\int x^4\sqrt{4+x^5}$

Do I make $u=4+x^5$ in order to solve this problem?

**mollymcf2009** · April 18th 2009, 09:46 PM

Originally Posted by cammywhite

$\int x^4\sqrt{4+x^5}$

Do I make $u=4+x^5$ in order to solve this problem?

Yes! Just remember to factor $\frac{1}{5}$ out of your integral!

**Prove It** · April 18th 2009, 09:46 PM

Originally Posted by cammywhite

$\int x^4\sqrt{4+x^5}$

Do I make $u=4+x^5$ in order to solve this problem?

Yes.

Notice that if $u = 4 + x^5$ then $\frac{du}{dx} = 5x^4$ .

So $\int{x^4\sqrt{4 + x^5}\,dx} = \frac{1}{5}\int{5x^4(4 + x^5)^{\frac{1}{2}}\,dx}$

$= \frac{1}{5}\int{u^{\frac{1}{2}}\,\frac{du}{dx}\,dx }$

$= \frac{1}{5}\int{u^{\frac{1}{2}}\,du}$ .

I trust you can go from here...

**cammywhite** · April 18th 2009, 10:14 PM

Originally Posted by Prove It

Yes.

Notice that if $u = 4 + x^5$ then $\frac{du}{dx} = 5x^4$ .

So $\int{x^4\sqrt{4 + x^5}\,dx} = \frac{1}{5}\int{5x^4(4 + x^5)^{\frac{1}{2}}\,dx}$

$= \frac{1}{5}\int{u^{\frac{1}{2}}\,\frac{du}{dx}\,dx }$

$= \frac{1}{5}\int{u^{\frac{1}{2}}\,du}$ .

I trust you can go from here...

Let me try

$= \frac{1}{5}\int{u^{\frac{1}{2}}\,du}$
$= \frac{1}{5}\int\frac{{u^{\frac{3}{2}}\,du}}{\frac{ 3}{2}}$

so far correct?

**mollymcf2009** · April 18th 2009, 10:24 PM

Originally Posted by cammywhite

Let me try

$= \frac{1}{5}\int{u^{\frac{1}{2}}\,du}$
$= \frac{1}{5}\int\frac{{u^{\frac{3}{2}}\,du}}{\frac{ 3}{2}}$

so far correct?

Your integration is correct, but you need to leave out the integral sign & du since you have already integrated.

**cammywhite** · April 18th 2009, 10:38 PM

Originally Posted by mollymcf2009

Your integration is correct, but you need to leave out the integral sign & du since you have already integrated.

I looked at the answer and it's $\frac {\sqrt{x^5+4}(2x^5+8)}{15}$

I have no clue how to get to the answer

**TheEmptySet** · April 18th 2009, 11:31 PM

Originally Posted by cammywhite

Let me try

$= \frac{1}{5}\int{u^{\frac{1}{2}}\,du}$
$= \frac{1}{5}\int\frac{{u^{\frac{3}{2}}\,du}}{\frac{ 3}{2}}$

so far correct?

$<br /> \frac{1}{5} \left( \frac{u^{3/2}}{\frac{3}{2}}\right)=\frac{2}{15}u^{3/2}<br />$

$\frac{2}{15}(4+x^5)^{3/2}=\frac{2}{15}\sqrt{4+x^5}(4+x^5)=\frac{2(4+x^5)\ sqrt{4+x^5}}{15}=\frac{(8+2x^5)\sqrt{4+x^5}}{15}$

**cammywhite** · April 18th 2009, 11:42 PM

Originally Posted by TheEmptySet

$<br /> \frac{1}{5} \left( \frac{u^{3/2}}{\frac{3}{2}}\right)=\frac{2}{15}u^{3/2}<br />$

$\frac{2}{15}(4+x^5)^{3/2}=\frac{2}{15}\sqrt{4+x^5}(4+x^5)=\frac{2(4+x^5)\ sqrt{4+x^5}}{15}=\frac{(8+2x^5)\sqrt{4+x^5}}{15}$

i don't get this part

$\frac{2}{15}(4+x^5)^{3/2} =\frac{2}{15}\sqrt{4+x^5}(4+x^5)$

**TheEmptySet** · April 18th 2009, 11:47 PM

Originally Posted by cammywhite

i don't get this part

$\frac{2}{15}(4+x^5)^{3/2} =\frac{2}{15}\sqrt{4+x^5}(4+x^5)$

We are using this exponent law

$x^ax^b=x^{a+b}$

Note that $x^{\frac{3}{2}}=x^{\frac{1}{2}}x^1=\sqrt{x}\cdot x$

So using this we get

$(4+x^5)^{3/2} =\sqrt{4+x^5}(4+x^5)$

**Scott H** · April 19th 2009, 08:10 AM

Integration by substitution is like a backward-application of the chain rule. In the expression

$x^4\sqrt{4+x^5},$

we see that $x^4$ looks like it was differentiated from $4+x^5$ of the square root.

To find the real antiderivative, we just add in the factor $\frac{1}{5}$ to get $x^4$ from $5x^4$ .

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