Integral 1 to 2 of (x^3)lnxdx
What do we use as u? I tried lnx and x^3 and it didnt seem to work.
You should remember the acronym LIATE: you want to choose for u whatever comes first in the list Logarithmic, Inverse trigonometric, Algebraic, Trigonometric, Exponential. In this case you would choose $\displaystyle u=\ln{x}$.
- Hollywood
Here's another method:
$\displaystyle \int_1^2 x^a dx=\frac{2^{1+a}-1}{1+a}$
Differentiate with respect to $\displaystyle a$:
$\displaystyle \int_1^2 x^a \log(x) dx=\frac{2^{1+a}\log(2)(1+a)-2^{1+a}+1}{(1+a)^2}$
Putting $\displaystyle a=3$, gives the answer
$\displaystyle \int_1^2 x^3 \log(x) dx= \frac{4\times 2^4 \log(2)-2^4+1}{4^2}=4\log(2)-\frac{15}{16}$