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<blockquote data-quote="bk12321" data-source="post: 3305961" data-attributes="member: 564687">Alright I need some more helpI did this problem already, answer at the bottom, if somebody could check my work id appreciate it, I just wanna make sure its right for class tomorrow.#6. Consider the indefinite integral: (integral) (4x^3+4x^2-27x-33)/(x^2+9) dxThe integrand decomposes into this form: ax + b + c/x-3 d/x+3Ive gotta find a,b,c, and d, and then integrate using partial fractions to get the antiderivative of the entire integral you see above.Using long division of x^2+9 into the big part, I get a remainder of 9x+3 and a quotient of 4x+4,so now I have 4x + 4 + (integral) (9x+3)/(x^2+9) dx. much simpler integral.partial fractions part:9x+3/(x+3)(x-3)therefore(A(x-3)+B(x+3))/((x+3)(x-3))again therefore((A+B)x+(-3A+3B))/(x+3)(x-3)using the equation I found from doing the long division... I knowA+B=9-3A+3B=3so 6b = 30 and B = 5A=4THEREFORE(integral) (4x+4) dx + (integral) (5/x-3) dx + (integral) (4/x+3) dx =2x^2 + 4x + (5)(ln(abs(x-3))) + (4)(ln(abs(x+3))) + Cis that right??? lol sorry I know its a pain to read typed out calc II</blockquote>

[QUOTE="bk12321, post: 3305961, member: 564687"] Alright I need some more help I did this problem already, answer at the bottom, if somebody could check my work id appreciate it, I just wanna make sure its right for class tomorrow. #6. Consider the indefinite integral: (integral) (4x^3+4x^2-27x-33)/(x^2+9) dx The integrand decomposes into this form: ax + b + c/x-3 d/x+3 Ive gotta find a,b,c, and d, and then integrate using partial fractions to get the antiderivative of the entire integral you see above. Using long division of x^2+9 into the big part, I get a remainder of 9x+3 and a quotient of 4x+4, so now I have 4x + 4 + (integral) (9x+3)/(x^2+9) dx. much simpler integral. partial fractions part: 9x+3/(x+3)(x-3) therefore (A(x-3)+B(x+3))/((x+3)(x-3)) again therefore ((A+B)x+(-3A+3B))/(x+3)(x-3) using the equation I found from doing the long division... I know A+B=9 -3A+3B=3 so 6b = 30 and B = 5 A=4 THEREFORE (integral) (4x+4) dx + (integral) (5/x-3) dx + (integral) (4/x+3) dx = 2x^2 + 4x + (5)(ln(abs(x-3))) + (4)(ln(abs(x+3))) + C is that right??? lol sorry I know its a pain to read typed out calc II [/QUOTE]

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