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Partial fractions + Integration

(4x^4-28x^3+51x^2-39x+1)/(2x-1)^2(x-5) dx the limits are 4 and 2. The answer in the book is 88/21 - 3/2ln3 +1/2ln7. I have tried this so many times and don't seem to be getting it right and sometimes the answers are wrong in my book, if someone got time can you try it out to confirm if the answer in the book is wrong:P..... The answer I got is ln7 - 2ln3 + 130/21
Reply 1
Original post by Merdan
(4x^4-28x^3+51x^2-39x+1)/(2x-1)^2(x-5) dx the limits are 4 and 2. The answer in the book is 88/21 - 3/2ln3 +1/2ln7. I have tried this so many times and don't seem to be getting it right and sometimes the answers are wrong in my book, if someone got time can you try it out to confirm if the answer in the book is wrong:P..... The answer I got is ln7 - 2ln3 + 130/21


http://www.wolframalpha.com/input/?i=integrate+%282%2C4%29+%284x%5E4-28x%5E3%2B51x%5E2-39x%2B1%29%2F%28%282x-1%29%5E2%28x-5%29%29
Reply 2
Original post by Merdan
(4x^4-28x^3+51x^2-39x+1)/(2x-1)^2(x-5) dx the limits are 4 and 2. The answer in the book is 88/21 - 3/2ln3 +1/2ln7. I have tried this so many times and don't seem to be getting it right and sometimes the answers are wrong in my book, if someone got time can you try it out to confirm if the answer in the book is wrong:P..... The answer I got is ln7 - 2ln3 + 130/21


The text book is right

1. With long division

4x428x3+51x239x+14x324x2+21x5=x1+6x213x4(2x1)2(x5)\displaystyle \frac{4x^4-28x^3+51x^2-39x+1}{4x^3-24x^2+21x-5}=x-1+\frac{6x^2-13x-4}{(2x-1)^2(x-5)}

2.
starting Partial fractions

A2x1+B(2x1)2+Cx5\displaystyle \frac{A}{2x-1}+\frac{B}{(2x-1)^2}+\frac{C}{x-5}

calculate A,B,C then integrate

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