<p>How do I do this integral?</p>
<p>1/[x*LN(x^3)] dx?</p>
<p>Thank you!</p>
<p>use a u-substitution with u = ln (x^3)</p>
<p>Actually no. What would du be? it would have to be (3x^2)/(x^3)dx and we dont have any extra x's.</p>
<p>TI89 says [LN(LN(x^3))]/3</p>
<p>Can't figure it out on paper yet... Ill get back to ya!</p>
<p>I don't think you can do it, because if you use u-sub, there's no extra term on the "outside" to balance the derivative of the inside function, just like what allanjyu said. i used mathematica and it spits the original integral right out, meaning, you can't do it. did you type it out right? what's the context of the problem, if there is one?</p>
<p>int[dx/(x*ln(x^3))]</p>
<p>Okay you use integration by parts.
Where: </p>
<p>u=ln(x^3)
du=[3x^2/x^3]dx = (3dx)/x
v=int[dx/x]=lnx.</p>
<p>uv-int[(v)(du)]=
(lnx)(ln(x^3)) - 3*int[lnxdx/x] THEN we can use classic u substitution for this second integral</p>
<p>u=lnx
du=dx/x</p>
<p>(lnx)(ln(x^3)) - 3<em>[int(udu)]=
(lnx)(ln(x^3)) - 3</em>[(u^2)/2]=</p>
<p>(lnx)(ln(x^3)) - 3*[(lnx^2)/2]</p>
<p>EDIT: SON OF A B**** I DID IT WRONG. the first u=1/ln(x^2) not u=ln(x^3)</p>
<p>CALCULUS GETS ME STEAMED!!!!</p>
<p>1/(xln(x^3)) can be rewritten as 1/(3x ln x), because ln x^n = n ln x</p>
<p>so you're left with the integral of 1/3(1/(x ln x))
u=ln x
du=1/x dx</p>
<p>And you get the right answer.</p>
<p>I think my method's much easier than allan's.</p>
<p>You get 1/3int[(1/x)(1/ln x)], which, after substitution, simplifies to:
1/3int[1/u du]</p>
<p>Mines definitely not right, i edited my post. AND HERE I WAS thinking oh man ill show those collegeconfidential nerds. BUT NO i failed.</p>
<p>i'm going to go watch some oc. The OC by definition is catharsis.</p>
<p>
</p>
<p>Mathematica definitely spits out an answer:</p>
<p>Integrate[ 1/(x*Log[x^3]), x] ==</p>
<p>Log[Log[x^3]]/3</p>
<p><a href="http://integrals.wolfram.com/%5B/url%5D">http://integrals.wolfram.com/</a></p>
<p>unless i'm reading the integrand incorrectly, you can do a u - sub. U = ln (x^3), thus du = 1/(x^3) * (3x^2) * dx. The x^2 and the x^3 cancel out forming du = 3/x * dx. You already have 1/x in the integral, so multiply it by 3 to get 3/x. Then just stick a 1/3 on the outside of the integral and you get 1/3 * integral of (du/u) which does come out to be</p>
<p>ln (ln (x^3))/3</p>
<p>pheonix and allan are both right, it wasn't that hard- easier if u used basic log rules to take out the power</p>