# help! integrals!

• Mar 5th 2008, 11:57 PM
s0urgrapes
help! integrals! - URGENT
integral of:
3(x^4/e^x^5)dx

I'm pretty sure its substitution rule but I don't know what to put as the u

u=x^4?
u=e^x^5 help!

and:

integral of:
5e^s/1+3e^s
• Mar 6th 2008, 12:40 AM
xifentoozlerix
${\text{Let }}u = e^{x^5 }\text{ ; }
du = 5x^4 e^{x^5 }\text{ } dx$

$\int {3\frac{{x^4 }}
{{e^{x^5 } }}} {\text{ }}dx = \frac{3}
{5}\int {\frac{{du}}
{{u^2 }}} = \frac{3}
{5}\left( { - \frac{1}
{u}} \right) = - \frac{3}
{{5u}} = - \frac{3}
{{5e^{x^5 } }}
= \boxed{ - \frac{{3e^{ - x^5 } }}
{5}}
$

${\text{Let }}u =1 + 3e^{s}\text{ ; }
du = 3e^{s}\text{ } ds$

$\int {\frac{{5e^s }}
{{1 + 3e^s }}} {\text{ }}ds = \frac{5}
{3}\int {\frac{{du}}
{u}} = \frac{5}
{3}\ln \left( {\left| u \right|} \right) = \frac{5}
{3}\ln \left( {\left| {1 + 3e^s } \right|} \right) = \frac{5}
{3}\ln \left( {1 + 3e^s } \right)

$
• Mar 6th 2008, 06:48 AM
s0urgrapes
Quote:

Originally Posted by xifentoozlerix
${\text{Let }}u = e^{x^5 }\text{ ; }
du = 5x^4 e^{x^5 }\text{ } dx$

$\int {3\frac{{x^4 }}
{{e^{x^5 } }}} {\text{ }}dx = \frac{3}
{5}\int {\frac{{du}}
{{u^2 }}} = \frac{3}
{5}\left( { - \frac{1}
{u}} \right) = - \frac{3}
{{5u}} = - \frac{3}
{{5e^{x^5 } }}
= \boxed{ - \frac{{3e^{ - x^5 } }}
{5}}
$

${\text{Let }}u =1 + 3e^{s}\text{ ; }
du = 3e^{s}\text{ } ds$

$\int {\frac{{5e^s }}
{{1 + 3e^s }}} {\text{ }}ds = \frac{5}
{3}\int {\frac{{du}}
{u}} = \frac{5}
{3}\ln \left( {\left| u \right|} \right) = \frac{5}
{3}\ln \left( {\left| {1 + 3e^s } \right|} \right) = \frac{5}
{3}\ln \left( {1 + 3e^s } \right)

$

thanks! just a question, how do you get du/u^2 in the first one, how come it isn't du/u
• Mar 6th 2008, 08:54 AM
xifentoozlerix
Well, let's see what happens if you do $\int {\frac{{du}}{u}}$:

$\int {\frac{{du}}
{u}} = \int {\frac{{5x^4 e^{x^5 } {\text{ }}dx}}
{{e^{x^5 } }}} = \int {5x^4 dx}
$

As you can see, the term ${e^{x^5 } }$ cancels and you end up with a different integral. With ${u^2 }$ however, the top ${e^{x^5 } }$ in the numerator is taken care of, and it leaves the ${e^{x^5 } }$ in the denominator as the original problem had:

$\int {\frac{{du}}
{{u^2 }}} = \int {\frac{{5x^4 e^{x^5 } {\text{ }}dx}}
{{\left( {e^{x^5 } } \right)^2 }}} = \int {\frac{{5x^4 dx}}
{{e^{x^5 } }}}$