How do you integrate ((3x2)−25x+43)dx(2x+1)((x−2)2)?
1 Answer
Explanation:
Lets rewrite integrand
3x2−25x+43(2x+1)(x−2)2=A2x+1+Bx−2+C(x−2)2=
=A(x−2)2+B(2x+1)(x−2)+C(2x+1)(2x+1)(x−2)2=
=A(x2−4x+4)+B(2x2−3x−2)+C(2x+1)(2x+1)(x−2)2=
=Ax2−4Ax+4A+2Bx2−3Bx−2B+2Cx+C(2x+1)(x−2)2=
=x2(A+2B)+x(−4A−3B+2C)+(4A−2B+C)(2x+1)(x−2)2
Comparing with integrand coefficients we get system of linear equations with 3 unknowns:
A+2B=3⇒A=3−2B
−4A−3B+2C=−25
4A−2B+C=43
From 1st eq insert into the 2nd and 3rd:
−4(3−2B)−3B+2C=−25
4(3−2B)−2B+C=43
−12+8B−3B+2C=−25
12−8B−2B+C=43
5B+2C=−13
−10B+C=31
5B+2C=−13
−20B+2C=62
25B=−75⇒B=−3
C=31+10B=31−30=1⇒C=1
A=3−2B=3+6=9⇒A=9
So, the integral is broken apart:
∫92x+1dx+∫−3x−2dx+∫1(x−2)2dx=I
2x+1=t,2dx=dt,dx=dt2,
x−2=u,dx=du,
x−2=m,dx=dm
I=∫9tdt2+∫−3udu+∫1m2dm=
=92∫dtt−3∫duu+∫m−2dm=
=92ln|t|−3ln|u|+m−1−1+C=
=92ln|2x+1|−3ln|x−2|−1x−2+C
