Quadratic Integral

To compute an integral of the form

[画像: int(dx)/(a+bx+cx^2), ]

(1)

complete the square in the denominator to obtain

[画像: int(dx)/(a+bx+cx^2)=1/cint(dx)/((x+b/(2c))^2+(a/c-(b^2)/(4c^2))). ]

(2)

Let u=x+b/2c. Then define

[画像: -A^2=a/c-(b^2)/(4c^2)=1/(4c^2)(4ac-b^2)=1/(4c^2)q, ]

(3)

where

q=4ac-b^2

(4)

is the negative of the polynomial discriminant. If q<0, then

[画像: A=1/(2c)sqrt(-q). ]

(5)

Now use partial fraction decomposition,

[画像: 1/cint(du)/((u+A)(u-A))=1/cint((A_1)/(u+A)+(A_2)/(u-A))du ]

(6)

[画像: ((A_1)/(u+A)+(A_2)/(u-A))=(A_1(u-A)+A_2(u+A))/(u^2-A^2) =((A_1+A_2)u+A(A_2-A_1))/(u^2-A^2), ]

(7)

so A_2+A_1=0=>A_2=-A_1 and A(A_2-A_1)=-2AA_1=1=>A_1=-1/(2A). Plugging these in,

[画像: 1/cint(-1/(2A)1/(u+A)+1/(2A)1/(u-A))du =1/(2Ac)[-ln(u+A)+ln(u-A)] =1/(2Ac)ln((u-A)/(u+A)) =1/(2(1/(2c))sqrt(-q)c)ln((x+b/(2c)-1/(2c)sqrt(-q))/(x+b/(2c)+1/(2c)sqrt(-q))) =1/(sqrt(-q))ln((2cx+b-sqrt(-q))/(2cx+b+sqrt(-q))) ]

(8)

for q<0. Note that this integral is also tabulated in Gradshteyn and Ryzhik (2000, equation 2.172), where it is given with a sign flipped.

See also

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References

Gradshteyn, I. S. and Ryzhik, I. M. Tables of Integrals, Series, and Products, 6th ed. San Diego, CA: Academic Press, 2000.

Referenced on Wolfram|Alpha

Quadratic Integral

Cite this as:

Weisstein, Eric W. "Quadratic Integral." From MathWorld--A Wolfram Resource. https://mathworld.wolfram.com/QuadraticIntegral.html

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