Evaluate the improper integral I=∫01xpln(x)dxI = \int_0^1 x^p \ln(x) dxI=∫01xpln(x)dx for p>−1p > -1p>−1.
−1(p+1)2-\frac{1}{(p+1)^2}−(p+1)21
1(p+1)2\frac{1}{(p+1)^2}(p+1)21
−1p+1-\frac{1}{p+1}−p+11
1p+1\frac{1}{p+1}p+11