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Chapter 7: Problem 20

Evaluate the following integrals. $$\int \frac{y+1}{y^{3}+3 y^{2}-18 y} d y$$

Short Answer

Expert verified
Question: Evaluate the integral: $$\int \frac{y+1}{y^{3}+3 y^{2}-18 y} dy$$ Answer: $$-\frac{1}{18}\ln|y| + \frac{7}{9}\ln|y+6| + \frac{4}{9}\ln|y-3| + C$$

Step by step solution

01

Factorize the denominator

Factor the denominator \(y^{3}+3 y^{2}-18 y\): $$y^{3}+3 y^{2}-18 y = y (y^{2}+ 3y -18) = y(y+6)(y-3)$$
02

Apply partial fraction decomposition

Express the integrand as partial fractions using the constants A, B, and C: $$\frac{y+1}{y (y+6)(y-3)} = \frac{A}{y} + \frac{B}{y+6} + \frac{C}{y-3}$$
03

Find the constants A, B, and C

To find A, B, and C, multiply both sides of the equation by the common denominator and simplify: $$(y+1) = A(y+6)(y-3)+ B(y)(y-3) + C(y)(y+6)$$ Now, substitute the suitable values of y to eliminate B and C so that we can solve for A: $$y=0: 1 = -18A \Rightarrow A = -\frac{1}{18}$$ $$y=-6: 7 = 9B \Rightarrow B = \frac{7}{9}$$ $$y=3: 4 = 9C \Rightarrow C = \frac{4}{9}$$ Now, substitute the values of A, B, and C back into the partial fractions: $$\frac{y+1}{y(y+6)(y-3)} = -\frac{1}{18}\frac{1}{y} + \frac{7}{9}\frac{1}{y+6} + \frac{4}{9}\frac{1}{y-3}$$
04

Integrate the partial fractions

Integrate each term with respect to y: $$\int \frac{y+1}{y(y+6)(y-3)} dy = \int\left(-\frac{1}{18}\frac{1}{y} + \frac{7}{9}\frac{1}{y+6} + \frac{4}{9}\frac{1}{y-3}\right) dy$$ $$=-\frac{1}{18}\int\frac{1}{y} dy + \frac{7}{9}\int\frac{1}{y+6} dy + \frac{4}{9}\int\frac{1}{y-3} dy$$ $$= -\frac{1}{18}\ln|y| + \frac{7}{9}\ln|y+6| + \frac{4}{9}\ln|y-3| + C$$ Therefore, the evaluated integral is: $$\int \frac{y+1}{y(y+6)(y-3)} dy = -\frac{1}{18}\ln|y| + \frac{7}{9}\ln|y+6| + \frac{4}{9}\ln|y-3| + C$$

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