Re: HSC 2013 2U Marathon
i
x^{a+1}ln(x) \ \ u= x^{a+1}\ \ u'=(a+1) x^a , \ \ v=ln(x) \ \ v'= \frac{1}{x}
ln(x)(a+1)x^a \ \ + \ \ \frac{1}{x} x^{(a+1)} = ln(x)(a+1)x^a + x^a
ii
\int_1^e \ \ x^{a} \ln x \ dx = \frac{1}{a+1} \big_((x^{(a+1)}\ \ ln(x))- \frac{x^{(a+1)}}{a+1})_1^e
$By...