Terms, terms and terms

Calculus Level 5

∫ − ∞ ∞ x 4 x 8 + 2 x 6 − 4 x 4 + 8 x 2 + 16 d x \large \int_{-\infty}^\infty \dfrac{x^4}{x^8 + 2x^6 - 4x^4 + 8x^2 + 16} \, dx

If the value of the integral above is equal to π a b − 12 c , \dfrac \pi a \sqrt{ \dfrac{b-\sqrt{12}}c }, , where a , b a,b and c c are positive integers greater than 1, find ( a + b + c ) 2 (a+b+c)^2 .


The answer is 7225.

Joel Yip by Joel Yip , 21, Singapore

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1 solution

Joel Yip
Feb 19, 2016

f ( x ) = x 4 x 8 + 2 x 6 − 4 x 4 + 8 x 2 + 16 = 1 x 4 + 2 x 2 − 4 + 8 x 2 + 16 x 4 = 1 ( x 2 + 4 x 2 ) 2 − 8 + 2 ( x 2 + 4 x 2 ) − 4 = 1 ( x 2 + 4 x 2 ) 2 + 2 ( x 2 + 4 x 2 ) − 12 = 1 ( x 2 + 4 x 2 ) 2 + 2 ( x 2 + 4 x 2 ) + 1 − 13 = 1 ( x 2 + 1 + 4 x 2 ) 2 − 13 \begin{aligned} f(x) & = \dfrac {x^4}{x^8+2x^6-4x^4+8x^2+16} \\ & = \dfrac{1}{x^4+2x^2-4+\frac{8}{x^2}+\frac{16}{x^4}} \\ & = \dfrac{1}{\left( x^2 + \frac{4}{x^2}\right)^2 - 8 + 2\left( x^2 + \frac{4}{x^2}\right) - 4} \\ & = \dfrac{1}{\left( x^2 + \frac{4}{x^2}\right)^2 + 2\left( x^2 + \frac{4}{x^2}\right) - 12} \\ & = \dfrac{1}{\left( x^2 + \frac{4}{x^2}\right)^2 + 2\left( x^2 + \frac{4}{x^2}\right) +1 - 13} \\ & = \dfrac{1}{\left( x^2 + 1 + \frac{4}{x^2}\right)^2 - 13} \end{aligned}

this one is from Chew-Seong Cheong.

1 ( x 2 + 1 + 4 x 2 ) 2 − 13 = 1 2 13 ( 1 x 2 + 1 + 4 x 2 + 13 − 1 x 2 + 1 + 4 x 2 − 13 ) \begin{matrix} \frac { 1 }{ { \left( { x }^{ 2 }+1+\frac { 4 }{ { x }^{ 2 } } \right) }^{ 2 }-13 } =\frac { 1 }{ 2\sqrt { 13 } } \left( \frac { 1 }{ { x }^{ 2 }+1+\frac { 4 }{ { x }^{ 2 } } +\sqrt { 13 } } -\frac { 1 }{ { x }^{ 2 }+1+\frac { 4 }{ { x }^{ 2 } } -\sqrt { 13 } } \right) \end{matrix}

∫ − ∞ ∞ 1 2 13 ( 1 x 2 + 1 + 4 x 2 + 13 − 1 x 2 + 1 + 4 x 2 − 13 ) d x = 1 2 13 ( ∫ − ∞ ∞ 1 x 2 + 1 + 4 x 2 + 13 d x − ∫ − ∞ ∞ 1 x 2 + 1 + 4 x 2 − 13 d x ) \int _{ -\infty }^{ \infty }{ \frac { 1 }{ 2\sqrt { 13 } } } \left( \frac { 1 }{ { x }^{ 2 }+1+\frac { 4 }{ { x }^{ 2 } } +\sqrt { 13 } } -\frac { 1 }{ { x }^{ 2 }+1+\frac { 4 }{ { x }^{ 2 } } -\sqrt { 13 } } \right) dx=\frac { 1 }{ 2\sqrt { 13 } } \left( \int _{ -\infty }^{ \infty }{ \frac { 1 }{ { x }^{ 2 }+1+\frac { 4 }{ { x }^{ 2 } } +\sqrt { 13 } } dx } -\int _{ -\infty }^{ \infty }{ \frac { 1 }{ { x }^{ 2 }+1+\frac { 4 }{ { x }^{ 2 } } -\sqrt { 13 } } dx } \right)

It took a lot of time to figure this out, 1 2 13 ( ∫ − ∞ ∞ 1 x 2 + 1 + 4 x 2 + 13 d x − ∫ − ∞ ∞ 1 x 2 + 1 + 4 x 2 − 13 d x ) = π 2 13 ( 5 + 13 12 − 5 − 13 12 ) \frac { 1 }{ 2\sqrt { 13 } } \left( \int _{ -\infty }^{ \infty }{ \frac { 1 }{ { x }^{ 2 }+1+\frac { 4 }{ { x }^{ 2 } } +\sqrt { 13 } } dx } -\int _{ -\infty }^{ \infty }{ \frac { 1 }{ { x }^{ 2 }+1+\frac { 4 }{ { x }^{ 2 } } -\sqrt { 13 } } dx } \right) =\frac { \pi }{ 2\sqrt { 13 } } \left( \sqrt { \frac { 5+\sqrt { 13 } }{ 12 } } -\sqrt { \frac { 5-\sqrt { 13 } }{ 12 } } \right)

π 2 13 ( 5 + 13 12 − 5 − 13 12 ) = π 2 13 ( 5 6 − 2 25 − 13 144 ) \frac { \pi }{ 2\sqrt { 13 } } \left( \sqrt { \frac { 5+\sqrt { 13 } }{ 12 } } -\sqrt { \frac { 5-\sqrt { 13 } }{ 12 } } \right) =\frac { \pi }{ 2\sqrt { 13 } } \left( \sqrt { \frac { 5 }{ 6 } -2\sqrt { \frac { 25-13 }{ 144 } } } \right)

π 2 13 ( 5 6 − 2 25 − 13 144 ) = π 2 13 ( 5 6 − 2 1 12 ) \frac { \pi }{ 2\sqrt { 13 } } \left( \sqrt { \frac { 5 }{ 6 } -2\sqrt { \frac { 25-13 }{ 144 } } } \right) =\frac { \pi }{ 2\sqrt { 13 } } \left( \sqrt { \frac { 5 }{ 6 } -2\sqrt { \frac { 1 }{ 12 } } } \right)

π 2 13 ( 5 6 − 2 1 12 ) = π 2 13 ( 5 − 12 6 ) ) \frac { \pi }{ 2\sqrt { 13 } } \left( \sqrt { \frac { 5 }{ 6 } -2\sqrt { \frac { 1 }{ 12 } } } \right) =\frac { \pi }{ 2\sqrt { 13 } } \left( \sqrt { \frac { 5-\sqrt { 12 } }{ 6 } } \right) )

π 2 13 ( 5 − 12 6 ) = π 2 ( 5 − 12 78 ) \frac { \pi }{ 2\sqrt { 13 } } \left( \sqrt { \frac { 5-\sqrt { 12 } }{ 6 } } \right) =\frac { \pi }{ 2 } \left( \sqrt { \frac { 5-\sqrt { 12 } }{ 78 } } \right)

so ( 5 + 78 + 2 ) 2 = 7225 { \left( 5+78+2 \right) }^{ 2 }=7225

1 Helpful 0 Interesting 0 Brilliant 0 Confused

Oh yeah!

∫ − ∞ ∞ 1 x 2 + 1 + 4 x 2 + y = i 2 π − y + ( y − 3 ) ( y + 5 ) − 1 + − y − ( y − 3 ) ( y + 5 ) − 1 \int _{ -\infty }^{ \infty }{ \frac { 1 }{ { x }^{ 2 }+1+\frac { 4 }{ { x }^{ 2 } } +y } } =\frac { i\sqrt { 2 } \pi }{ \sqrt { -y+\sqrt { \left( y-3 \right) \left( y+5 \right) } -1 } +\sqrt { -y-\sqrt { \left( y-3 \right) \left( y+5 \right) } -1 } }

so, ∫ − ∞ ∞ 1 x 2 + 1 + 4 x 2 − y = i 2 π y + ( − y − 3 ) ( − y + 5 ) − 1 + y − ( − y − 3 ) ( − y + 5 ) − 1 \int _{ -\infty }^{ \infty }{ \frac { 1 }{ { x }^{ 2 }+1+\frac { 4 }{ { x }^{ 2 } } -y } } =\frac { i\sqrt { 2 } \pi }{ \sqrt { y+\sqrt { \left( -y-3 \right) \left( -y+5 \right) } -1 } +\sqrt { y-\sqrt { \left( -y-3 \right) \left( -y+5 \right) } -1 } }

∫ − ∞ ∞ 1 x 2 + 1 + 4 x 2 − y − 1 x 2 + 1 + 4 x 2 + y = i 2 π y + ( − y − 3 ) ( − y + 5 ) − 1 + y − ( − y − 3 ) ( − y + 5 ) − 1 − i 2 π − y + ( y − 3 ) ( y + 5 ) − 1 + − y − ( y − 3 ) ( y + 5 ) − 1 \int _{ -\infty }^{ \infty }{ \frac { 1 }{ { x }^{ 2 }+1+\frac { 4 }{ { x }^{ 2 } } -y } -\frac { 1 }{ { x }^{ 2 }+1+\frac { 4 }{ { x }^{ 2 } } +y } } =\frac { i\sqrt { 2 } \pi }{ \sqrt { y+\sqrt { \left( -y-3 \right) \left( -y+5 \right) } -1 } +\sqrt { y-\sqrt { \left( -y-3 \right) \left( -y+5 \right) } -1 } } -\frac { i\sqrt { 2 } \pi }{ \sqrt { -y+\sqrt { \left( y-3 \right) \left( y+5 \right) } -1 } +\sqrt { -y-\sqrt { \left( y-3 \right) \left( y+5 \right) } -1 } }

AND, ∫ − ∞ ∞ 1 ( x 2 + 1 + 4 x 2 ) 2 − y 2 = 1 2 y ( i 2 π y + ( − y − 3 ) ( − y + 5 ) − 1 + y − ( − y − 3 ) ( − y + 5 ) − 1 − i 2 π − y + ( y − 3 ) ( y + 5 ) − 1 + − y − ( y − 3 ) ( y + 5 ) − 1 ) \int _{ -\infty }^{ \infty }{ \frac { 1 }{ { \left( { x }^{ 2 }+1+\frac { 4 }{ { x }^{ 2 } } \right) }^{ 2 }-{ y }^{ 2 } } } =\frac { 1 }{ 2y } \left( \frac { i\sqrt { 2 } \pi }{ \sqrt { y+\sqrt { \left( -y-3 \right) \left( -y+5 \right) } -1 } +\sqrt { y-\sqrt { \left( -y-3 \right) \left( -y+5 \right) } -1 } } -\frac { i\sqrt { 2 } \pi }{ \sqrt { -y+\sqrt { \left( y-3 \right) \left( y+5 \right) } -1 } +\sqrt { -y-\sqrt { \left( y-3 \right) \left( y+5 \right) } -1 } } \right)

If y = 13 y=\sqrt { 13 } ,

∫ − ∞ ∞ 1 ( x 2 + 1 + 4 x 2 ) 2 − 13 = 1 2 13 ( i 2 π 13 + ( − 13 − 3 ) ( − 13 + 5 ) − 1 + 13 − ( − 13 − 3 ) ( − 13 + 5 ) − 1 − i 2 π − 13 + ( 13 − 3 ) ( 13 + 5 ) − 1 + − 13 − ( 13 − 3 ) ( 13 + 5 ) − 1 ) \int _{ -\infty }^{ \infty }{ \frac { 1 }{ { \left( { x }^{ 2 }+1+\frac { 4 }{ { x }^{ 2 } } \right) }^{ 2 }-13 } } =\frac { 1 }{ 2\sqrt { 13 } } \left( \frac { i\sqrt { 2 } \pi }{ \sqrt { \sqrt { 13 } +\sqrt { \left( -\sqrt { 13 } -3 \right) \left( -\sqrt { 13 } +5 \right) } -1 } +\sqrt { \sqrt { 13 } -\sqrt { \left( -\sqrt { 13 } -3 \right) \left( -\sqrt { 13 } +5 \right) } -1 } } -\frac { i\sqrt { 2 } \pi }{ \sqrt { -\sqrt { 13 } +\sqrt { \left( \sqrt { 13 } -3 \right) \left( \sqrt { 13 } +5 \right) } -1 } +\sqrt { -\sqrt { 13 } -\sqrt { \left( \sqrt { 13 } -3 \right) \left( \sqrt { 13 } +5 \right) } -1 } } \right)

1 2 13 ( i 2 π 13 + ( − 13 − 3 ) ( − 13 + 5 ) − 1 + 13 − ( − 13 − 3 ) ( − 13 + 5 ) − 1 − i 2 π − 13 + ( 13 − 3 ) ( 13 + 5 ) − 1 + − 13 − ( 13 − 3 ) ( 13 + 5 ) − 1 ) = π 2 5 + 12 78 \frac { 1 }{ 2\sqrt { 13 } } \left( \frac { i\sqrt { 2 } \pi }{ \sqrt { \sqrt { 13 } +\sqrt { \left( -\sqrt { 13 } -3 \right) \left( -\sqrt { 13 } +5 \right) } -1 } +\sqrt { \sqrt { 13 } -\sqrt { \left( -\sqrt { 13 } -3 \right) \left( -\sqrt { 13 } +5 \right) } -1 } } -\frac { i\sqrt { 2 } \pi }{ \sqrt { -\sqrt { 13 } +\sqrt { \left( \sqrt { 13 } -3 \right) \left( \sqrt { 13 } +5 \right) } -1 } +\sqrt { -\sqrt { 13 } -\sqrt { \left( \sqrt { 13 } -3 \right) \left( \sqrt { 13 } +5 \right) } -1 } } \right) =\frac { \pi }{ 2 } \sqrt { \frac { 5+\sqrt { 12 } }{ 78 } }

Joel Yip - 5 years, 3 months ago

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