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To calculate : ∫x2−5x+16(2x+1)(x−2)2dx
Answer to Problem 20E
3ln(|2x+1|)2−2x−2−ln(|x−2|)+C
Explanation of Solution
Given information :
∫x2−5x+16(2x+1)(x−2)2dx
Formula used :
Partial fraction decomposition:
px2+qx+r(x−a)2(x−b)=Ax−a+B(x−a)2+Cx−b
∫1x dx=ln(x)
Power rule: ∫andx=an+1n+1, n≠−1
Calculation:
∫x2−5x+16(2x+1)(x−2)2dx
Apply partial fraction decomposition:
x2−5x+16(2x+1)(x−2)2=A(x−2)+B(x−2)2+C(2x+1)
Write the right-hand side as a single fraction:
x2−5x+16(2x+1)(x−2)2=A(x−2)(2x+1)+B(2x+1)+(x−2)2C(2x+1)
The denominators are equal, so we require the equality of the numerators:
x2−5x+16=A(x−2)(2x+1)+B(2x+1)+(x−2)2C
Expand the right-hand side:
x2−5x+16=2x2A+x2C−3xA+2xB−4xC−2A+B+4C
Collect up the like terms:
x2−5x+16=x2(2A+C)+x(−3A+2B−4C)−2A+B+4C
The coefficients near the like terms should be equal, so the following system is obtained:
x2−5x+16=x2(2A+C)+x(−3A+2B−4C)−2A+B+4C{2A+C=1 ....(1)−3A+2B−4C=−5 ....(2)−2A+B+4C=16 ....(3)
Solve (1), (2) and (3) , we get
A=−1, B=2, C=3
Therefore,
x2−5x+16(2x+1)(x−2)2=−1(x−2)+2(x−2)2+3(2x+1)
⇒∫x2−5x+16(2x+1)(x−2)2dx=∫(−1(x−2)+2(x−2)2+3(2x+1))dx
=2∫1(x−2)2dx−∫1(x−2)dx+3∫1(2x+1)dx
Apply substitution: u=x−2⇒dudx=1⇒dx=duApply substitution: v=x−2⇒dvdx=1⇒dx=dvApply substitution: w=2x+1⇒dwdx=2⇒dx=12dw
=2∫1(u)2du−∫1vdv+3∫1w×12dw
=2∫u−2du−∫1vdv+32∫1wdw (∵a−n=1an)
=2[u−2+1−2+1]−ln(v)+3ln(w)2 (∵Power rule)
=−2u−ln(v)+3ln(w)2 (∵a−n=1an)
=3ln(w)2−2u−ln(v)
Substitute back: u=x−2, v=x−2, w=2x+1⇒∫x2−5x+16(2x+1)(x−2)2dx=3ln(|2x+1|)2−2x−2−ln(|x−2|)+C
Chapter G Solutions
Single Variable Calculus: Concepts and Contexts, Enhanced Edition
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