Question

integral[(lnx)^ndx] = x(lnx)^n - n integral[(lnx)^(n-1)dx
Use the above reduction formula to calculate the integral from [1,e], (ln(2x))^3 dx.

Answer 1

\[\int_1^e (\ln2x)^3~dx\]
Substitute \(u=2x\), so \(\dfrac{1}{2}du=dx\). As \(x\to1\), you have \(u\to2\), and as \(x\to e\), you have \(u\to2e\).
\[\frac{1}{2}\int_2^{2e}(\ln u)^3~du\]
By the reduction formula, you have that this expression is equivalent to
\[\frac{1}{2}\left(\left[u(\ln u)^3\right]_2^{2e}-3\int_2^{2e}(\ln u)^2~du\right)\]
Applying the formula again, you get
\[\frac{1}{2}\left(\left[u(\ln u)^3\right]_2^{2e}-3\left(\left[u(\ln u)^2\right]_2^{2e}-2\int_2^{2e}\ln u~du\right)\right)\]
And one more time:
\[\frac{1}{2}\left(\left[u(\ln u)^3\right]_2^{2e}-3\left(\left[u(\ln u)^2\right]_2^{2e}-2\left(\left[u\ln u\right]_2^{2e}-\int_2^{2e}du\right)\right)\right)\]
Hope I didn't make any errors...