Question

[(ax^2-bx)/(x+b)]dx怎么积分呢?

Answer 1

+x=t,dx=dt,x=t-b

[(ax^2-bx)/(x+b)]dx= [(a(t-b)^2-b(t-b))/tdt

=[at-(2ab+b)+(ab^2+b^2)/t]dt

=at^2/2-(2ab+b)t+(ab^2+b^2)lnt

=a(x+b)^2/2-(2ab+b)(x+b)+(ab^2+b^2)ln(x+b)