Question #898f1

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Answer 1 · 2017-05-04T00:33:12

Factoring the denominator using the pattern:

$(a^{3} - b^{3}) = (a - b) (a^{2} b^{0} + a^{1} b^{1} + a^{0} b^{2})$ $(a^3-b^3) = (a-b)(a^2b^0+a^1b^1+a^0b^2)$

$:.$

$(x^3-27)=(x-3)(x^2+3x+9)$

Decompose the fraction:

$(5x^2 -2x +42)/(x^3 -27) = A/(x-3)+ (Bx)/(x^2+3x+9)+ C/(x^2+3x+9)$

Multiply both sides by the denominator:

$5x^2 -2x +42 = A(x^2+3x+9)+ (Bx)(x-3)+ C(x-3)$

Let x = 3

$81= 27A+ 0B + 0B$

$A = 3$

$5x^2 -2x +42 = 3(x^2+3x+9)+ (Bx)(x-3)+ C(x-3)$

Let x = 0

$42 = 3(9)+ C(-3)$

$-3C=15$

$C = -5$

$5x^2 -2x +42 = 3(x^2+3x+9)+ (Bx)(x-3)-5(x-3)$

Let x = 1

$5 -2 +42 = 3(1+3+9)+ (B)(-2)-5(-2)$

$-4 = -2B$

$B = 2$

$(5x^2 -2x +42)/(x^3 -27) = 3/(x-3)+ (2x)/(x^2+3x+9)- 5/(x^2+3x+9)$

I checked this on a scratchpad.

Add Integral symbols:

$int(5x^2 -2x +42)/(x^3 -27)dx = 3int1/(x-3)dx+ int(2x)/(x^2+3x+9)dx- 5int1/(x^2+3x+9)dx$

Add 3 to the second integral and subtract 3 from the third:

$int(5x^2 -2x +42)/(x^3 -27)dx = 3int1/(x-3)dx+ int(2x+3)/(x^2+3x+9)dx- 8int1/(x^2+3x+9)dx$

The first two integrals become natural logarithms:

$int(5x^2 -2x +42)/(x^3 -27)dx = 3ln|x-3|+ ln(x^2+3x+9)- 8int1/(x^2+3x+9)dx$

The third integral is done by completing the square and a trigonometric substitution:

$int(5x^2 -2x +42)/(x^3 -27)dx = 3ln|x-3|+ ln(x^2+3x+9)- (16sqrt(3))/9tan^-1(sqrt3/9(2x+2))+ C$