Đặt u=ln(x2+x+1) ;dv=dx(x+1)2
du=2x+1x2+x+1dx ;v=−1x+1
I=−ln(x2+x+1)1x+1|10+1∫02x+1(x2+x+1)(x+1)dx
I=−12ln3+J
Với J
J=1∫02x+1(x2+x+1)(x+1)dx
J=1∫0x+2(x2+x+1)dx−1∫01(x+1)dx
J=1∫0x+1/2(x2+x+1)dx+1∫03/2(x2+x+1)dx−1∫01(x+1)dx
J=121∫0d(2x+1)(x2+x+1)dx+321∫01((x+1/2)2+3/4)dx−1∫01(x+1)dx
J=ln(x2+x+1)|10+J1−ln|x+1||10=
J=ln(3)+32J1−ln2
Với J2=∫1((x+1/2)2+3/4)dx (chút nữa thay cận vào là xong)
Đặt x+1/2=√32tant→dx=√321cos2tdt=√32(1+tan2t)dt
J2=∫√3/2(1+tan2t)3/4(1+tan2t)dt=2√3t+C=2√3arctan(x+1/2)+C
J1=2√3arctan(x+1/2)|10=2√3(arctan(3/2)−arctan(1/2))
Vậy I=−12ln3+ln3−ln2+322√3(arctan(3/2)−arctan(1/2))
hay
I=12ln3−ln2+√3(arctan(3/2)−arctan(1/2))
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