Question

∫（x^2）dx/(x+1)(x+3) 的不定积分怎么求?

Answer 1

∫1/(X+1）（x+3） dx

=∫dx/2(x+1)-∫dx/2(x+3)

=1/2*∫d(x+1)/(x+1)-1/2*∫d(x+3)/(x+3)

=ln(x+1)/2-ln(x+3)/2+C

=1/2*[ln(x+1)-ln(x+3)]+C

=ln[(x+1)/(x+3)]/2+C 注意分母2不在ln内,在ln外