El Niño integral

Calculus Level 5

∫ 1 2 x 2 x 2 + 5 x + 7 d x = A B ln ⁡ ( C D ) + E tan ⁡ − 1 ( F G ) H I − J tan ⁡ − 1 ( K L ) M N \int_{1}^{2} \dfrac{x}{2x^2+5x+7} \, dx = \frac{A}{B} \ln\left(\frac{C}{D}\right)+\frac{E\tan^{-1}\left(\frac{F}{\sqrt{G}}\right)}{H\sqrt{I}} - \frac{J\tan^{-1}\left(\frac{K}{\sqrt{L}} \right) }{M\sqrt{N}}

The equation above holds true for positive integers A , B , C , D , E , F , G , H , I , J , K , L , M A,B,C,D,E,F,G,H,I,J,K,L,M and N N such that gcd ⁡ ( A , B ) = gcd ⁡ ( C , D ) = gcd ⁡ ( E , F ) = gcd ⁡ ( K , L ) = gcd ⁡ ( J , M ) = 1 , \gcd(A,B) = \gcd(C,D) = \gcd(E,F) = \gcd(K,L) = \gcd(J,M ) = 1, with I I and N N square-free.

Find A + B + C + D + E + F + G + H + I + J + K + L + M + N A+B+C+D+E+F+G+H+I+J+K+L+M+N .


The answer is 204.

A Former Brilliant Member by A Former Brilliant Member

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1 solution

Chew-Seong Cheong
Mar 21, 2018

I = ∫ 1 2 x 2 x 2 + 5 x + 7 d x = 1 4 ∫ 1 2 4 x + 5 − 5 2 x 2 + 5 x + 7 d x = 1 4 ln ⁡ ( 2 x 2 + 5 x + 7 ) ∣ 1 2 − 5 8 ∫ 1 2 d x x 2 + 5 2 x + 7 2 = 1 4 ln ⁡ ( 25 14 ) − 5 8 ∫ 1 2 d x ( x + 5 4 ) 2 + 31 16 = 1 4 ln ⁡ ( 25 14 ) − 5 8 ⋅ 31 4 tan ⁡ − 1 ( x + 5 4 31 4 ) ∣ 1 2 = 1 4 ln ⁡ ( 25 14 ) − 5 2 31 tan ⁡ − 1 ( 4 x + 5 31 ) ∣ 1 2 = 1 4 ln ⁡ ( 25 14 ) + 5 2 31 [ tan ⁡ − 1 ( 9 31 ) − tan ⁡ − 1 ( 13 31 ) ] \begin{aligned} I & = \int_1^2 \frac x{2x^2+5x+7}dx \\ & = \frac 14 \int_1^2 \frac {4x+5-5}{2x^2+5x+7}dx \\ & = \frac 14 \ln(2x^2+5x+7)\bigg|_1^2 - \frac 58\int_1^2 \frac {dx}{x^2+\frac 52x + \frac 72} \\ & = \frac 14 \ln\left(\frac {25}{14}\right) - \frac 58\int_1^2 \frac {dx}{\left(x+\frac 54\right)^2 + \frac {31}{16}} \\ & = \frac 14 \ln\left(\frac {25}{14}\right) - \frac 5{8 \cdot \frac {\sqrt{31}}4} \tan^{-1}\left(\frac {x + \frac 54}{\frac {\sqrt{31}}4} \right)\bigg|_1^2 \\ & = \frac 14 \ln\left(\frac {25}{14}\right) - \frac 5{2\sqrt{31}} \tan^{-1}\left(\frac {4x + 5}{\sqrt{31}} \right)\bigg|_1^2 \\ & = \frac 14 \ln\left(\frac {25}{14}\right) + \frac 5{2\sqrt{31}} \left[\tan^{-1}\left(\frac 9{\sqrt{31}}\right) - \tan^{-1}\left(\frac {13}{\sqrt{31}}\right) \right] \end{aligned}

Therefore, A + B + C + D + E + F + G + H + I + J + K + L + M + N = 1 + 4 + 25 + 14 + 5 + 9 + 31 + 2 + 31 + 5 + 13 + 31 + 2 + 31 = 204 A+B+C+D+E+F+G+H+I+J+K+L+M+N=1+4+25+14+5+9+31+2+31+5+13+31+2+31=\boxed{204} .

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