# Two Integrals

• May 8th 2008, 06:27 AM
Hibijibi
Two Integrals
1) x^3*(x^2-1)^(1/3) dx

2) 1/(xln(x^3)) dx

Appreciate all help given. Thank you.
• May 8th 2008, 06:30 AM
PaulRS
2. Let $u=\ln(x)$ and remember that: $\ln(x^3)=3\cdot{\ln(x)}$
1. Note that: $
x^3 \cdot \left( {x^2 - 1} \right)^{\tfrac{1}
{3}} = x \cdot \left( {x^2 - 1 + 1} \right) \cdot \left( {x^2 - 1} \right)^{\tfrac{1}
{3}} = x \cdot \left( {x^2 - 1} \right)^{\tfrac{1}
{3} + 1} + x \cdot \left( {x^2 - 1} \right)^{\tfrac{1}
{3}}
$

and now try a substitution
• May 8th 2008, 06:35 AM
Moo
Hello,

Quote:

Originally Posted by Hibijibi
1) x^3*(x^2-1)^(1/3) dx

$=x^2 \cdot x \cdot (x^2-1)^{\frac 13}$

Integrate by parts :

$u(x)=x^2$

\begin{aligned} v'(x) &=x \cdot (x^2-1)^{\frac 13} \\
&=\frac 12 \cdot 2x \cdot (x^2-1)^{\frac 13} \end{aligned}

\Longrightarrow \begin{aligned} v(x) &=\frac 12 \cdot \frac{(x^2-1)^{\frac 43}}{\frac 43} \\
&=\frac 38 \cdot (x^2-1)^{\frac 43} \end{aligned}

etc...
• May 8th 2008, 08:10 AM
Hibijibi
So I rewrite the ln function as 1/ (x (3*ln(x))) and use u to equal lnx, and u'= 1/x, and then what?

Edit: i got the function down to 1/3 intergral of 1/u du, and how do I get past this?
• May 8th 2008, 08:11 AM
Moo
Quote:

Originally Posted by Hibijibi
So I rewrite the ln function as 1/ (x (3*ln(x))) and use u to equal lnx, and u'= 1/x, and then what?

$u=\ln x$

-> $du=\frac{dx}{x}$

$dx=x du$

do the substitutions :)
• May 8th 2008, 08:33 AM
Hibijibi
Thanks, I got it. Except I I used du = 1/x dx, since 1/x dx is in the function already and I solved. Appreciate it!
• May 8th 2008, 08:36 AM
Moo
Quote:

Originally Posted by Hibijibi
Thanks, I got it. Except I I used du = 1/x dx, since 1/x dx is in the function already and I solved. Appreciate it!

I'm sorry about it... Actually I'm a bit lost concerning the way people learn how to do substitutions :)
• May 8th 2008, 08:37 AM
Hibijibi
Haha everyone to their own right? :) Thank you very much Moo! *hugs*