1. ## Two Integrals

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

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

Appreciate all help given. Thank you.

2. 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

3. Hello,

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...

4. 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?

5. 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

6. 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!

7. 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

8. Haha everyone to their own right? Thank you very much Moo! *hugs*