Integration Mania Continued v2

Calculus Level 5

∫ 2 3 2 x 5 + x 4 − 2 x 3 + 2 x 2 + 1 x 6 + x 4 − x 2 − 1 d x \large \int_2^3 \dfrac{2x^5 + x^4 - 2x^3 + 2x^2 + 1}{x^6 + x^4-x^2 - 1} \, dx

If the integral above can be expressed in the form of

A B ln ⁡ 6 − D E , \dfrac AB \ln 6 - \dfrac DE \; ,

where A , B , D A,B,D and E E are positive integers with gcd ⁡ ( A , B ) = gcd ⁡ ( D , E ) = 1 \gcd(A,B) = \gcd(D,E) = 1 , find 6 × A × B × D × E 6\times A \times B \times D \times E .


The answer is 120.

Kunal Gupta by Kunal Gupta , 23, India

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1 solution

First Last
Apr 2, 2016

Denominator can be factored to ( x 2 + 1 ) 2 ( x 2 − 1 ) (x^2+1)^2(x^2-1) . Breaking up the fraction to ∫ 2 3 2 x 5 − 2 x 3 ( x 2 + 1 ) 2 ( x 2 − 1 ) d x + x 4 + 2 x 2 + 1 ( x 2 + 1 ) 2 ( x 2 − 1 ) d x \int_{2}^{3}\dfrac{2x^5-2x^3}{(x^2+1)^2(x^2-1)}dx + \dfrac{x^4+2x^2+1}{(x^2+1)^2(x^2-1)}dx

Canceling leaves ∫ 2 3 2 x 3 ( x 2 + 1 ) 2 d x + 1 ( x 2 − 1 ) d x = \int_{2}^{3}\dfrac{2x^3}{(x^2+1)^2}dx + \dfrac{1}{(x^2-1)}dx =

[ − x 2 x 2 + 1 + ln ⁡ ( x 2 + 1 ) ] 2 3 + [ ln ⁡ ( x − 1 x + 1 ) 2 ] 2 3 = \Bigg[\dfrac{-x^2}{x^2+1} + \ln(x^2+1)\Bigg]_{2}^{3} + \Bigg[\dfrac{\ln(\frac{x-1}{x+1})}{2}\Bigg]_{2}^{3} =

ln ⁡ ( 6 ) 2 − 1 10 \boxed{\frac{\ln(6)}{2} - \frac{1}{10}}

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