174x Filetype PDF File size 0.06 MB Source: wrean.ca
28.10 #20, from Basic Technical Mathematics with Calculus (8th Edition) by Allyn J. Washington. Z 3 2 −x +x +x+3 2 2 dx (x+1)(x +1) This example involves a repeated irreducible quadratic factor. The partial fractions set-up is 3 2 −x +x +x+3 = A +Bx+C+ Dx+E . 2 2 2 2 2 (x+1)(x +1) x+1 x +1 (x +1) 2 2 Multiplying both sides by (x + 1)(x +1) we obtain 3 2 2 2 2 −x +x +x+3=A(x +1) +(Bx+C)(x+1)(x +1)+(Dx+E)(x+1). (1) For x = −1, we get: 4=4A+0+0 =⇒ A=1. To obtain B, C, D, and E we will expand equation (1) and equate coefficients of powers of x. 3 2 4 2 3 2 −x +x +x+3=A(x +2x +1)+(Bx+C)(x +x +x+1)+(Dx+E)(x+1) 4 3 2 =(A+B)x +(B+C)x +(2A+B+C+D)x +(B+C+D+E)x+(A+C+E) By equating the corresponding coefficients of powers of x, we obtain A+B=0 B+C=−1 2A+B+C+D=1 B+C+D+E=1 A+C+E=3. Since A = 1, we can deduce B, C, D, and E as follows. A+B=0 =⇒ B=−1 B+C=−1 =⇒ C=0 A+C+E=3 =⇒ E=2 B+C+D+E=1 =⇒ D=0 Therefore, Z Z 3 2 −x +x +x+3dx= 1 − x + 2 dx. 2 2 2 2 2 (x+1)(x +1) x+1 x +1 (x +1) Wecanusethesubstitution u = x2 +1 to get Z x 1 2 x2 +1 dx = 2 ln(x +1)+C. Wecanusethetrigonometric substitution x = tanθ to deduce that Z 2 2 2 dx = 2x +tan−1x+C. (x +1) x +1 Wethenconclude that Z 3 2 −x +x +x+3 1 2 x −1 2 2 dx = ln|x+1|− ln(x +1)+ 2 +tan x+C. (x+1)(x +1) 2 x +1 Gilles Cazelais. Typeset with LAT X on February 18, 2008. E
no reviews yet
Please Login to review.