1) x^3*(x^2-1)^(1/3) dx

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

Appreciate all help given. Thank you.

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- May 8th 2008, 06:27 AMHibijibiTwo 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 AMPaulRS
2. Let $\displaystyle u=\ln(x)$ and remember that: $\displaystyle \ln(x^3)=3\cdot{\ln(x)}$

1. Note that: $\displaystyle

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 AMMoo
Hello,

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

Integrate by parts :

$\displaystyle u(x)=x^2$

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

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

$\displaystyle \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 AMHibijibi
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 AMMoo
- May 8th 2008, 08:33 AMHibijibi
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 AMMoo
- May 8th 2008, 08:37 AMHibijibi
Haha everyone to their own right? :) Thank you very much Moo! *hugs*