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Mathematics Dictionary
Dr. K. G. Shih
Integral : Rational and irrational functions
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CA 16 01. Basic formula from anti-derivative

Integrals of ration function from anti-differentiation
  • ∫(1/(1 + x^2))dx = arctan(x)
  • ∫(1/(1 - x^2))dx = arctanh(x)
Integrals of ir-ration function from anti-differentiation
  • ∫(1/Sqr(1 - x^2))dx = arcsin(x)
  • ∫(1/Sqr(1 + x^2))dx = arcsinh(x)
  • ∫(1/(x*Sqr(x^2 - 1))dx = arcsec(x)
  • ∫(1/(x*Sqr(1 - x^2))dx = arcsech(x)

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CA 16 02. ∫(1/(a + b*x^2))dx = Sqr(1/a*b)*arctan(sqr(b/a)*x)

  • 1/(a + b*x^2) = (1/a)/(1 + (b/a)*(x^2))
  • Let Sqr(b/a)*x = u, then du =Sqr(b/a)dx
  • Hence integral = ∫(1/Sqr(b/a))*((1/a)/(1 + u^2)))du
  • = (1/(Sqr(b/a)*a))*arctan(u)
  • = (Sqr(a/b)/a)*arctan(Sqr(b/a)*x) + c
  • = Sqr(1/a*b)*arctan(Sqr(b/a)*x) + c

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CA 16 03. ∫(1/Sqr(a - b*x^2))dx = Sqr(1/a*b)*arcsin(sqr(b/a)*x)

  • 1/Sqr(a - b*x^2) = (1/a)/(1 - (b/a)*(x^2))
  • Let Sqr(b/a)*x = u, then du =Sqr(b/a)dx
  • Hence integral = ∫(1/Sqr(b/a))*((1/a)/Sqr(1 - u^2)))du
  • = (1/(Sqr(b/a)*a))*arcsin(u)
  • = (Sqr(a/b)/a)*arcsin(Sqr(b/a)*x) + c
  • = Sqr(1/a*b)*arcsin(Sqr(b/a)*x) + c

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CA 16 04. ∫(1/(a*x^2 + b*x + c))dx = A/(x+B) : case 1

Case 1 : b^2 - 4*a*c = 0
  • ∫(1/(a*x^2 + b*x + c))dx = A/(x + B)
  • Find A and B
    • Using completing the square
    • (a*x^2 + b*x + c) = a*(x^2 + b*x/a + (b/(2*a))^2 - (b/(2*a))^2 + c/a)
    • = a*(x + b/(2*a))^2 - (b^2 - 4*a*c)/(4*a)
    • Since b^2 - 4*a*c = 0
    • Hence (a*x^2 + b*x + c)= a*(x + b/(2*a))^2 since b^2 - 4*a*c = 0
    • x + b/(2*a) = u and du = dx
    • ∫(1/(a*x^2 + b*x + c))dx
    • ∫ = (1/a)*(1/u^2)du
    • = (1/a)*(-2+1)*u^(-1)
    • = -1/(x + b/(2*a))
    • Hence A = -1/a and B = b/(2*a)
Example : Find ∫(1/(x^2 + 4*x + 4))dx
  • Since x^2 + 4*x + 4 = (x + 2)^2
  • Hence ∫(1/(x^2 + 4*x + 4))dx ∫(1/(x + 2)^2)dx
  • = -1/(x + 2)
b^2 - 4*a*c GT 0
  • ∫(1/(a*x^2 + b*x + c))dx = A*ln(x + B)+C*ln(x - B)
  • Find A, B and C

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CA 16 05. ∫(1/(a*x^2 + b*x + c))dx = A*arctan(B*(x+C)) : case 2

Case 2 : b^2 - 4*a*c LT 0
  • ∫(1/(a*x^2 + b*x + c))dx = A*arctan(B*(x + C))
  • Find A and B
    • Using completing the square
    • (a*x^2 + b*x + c) = a*(x^2 + b*x/a + (b/(2*a))^2 - (b/(2*a))^2 + c/a)
    • = a*(x + b/(2*a))^2 - (b^2 - 4*a*c)/(4*a)
    • Since b^2 - 4*a*c LT 0
    • Hence (a*x^2 + b*x + c)= a*(x + b/(2*a))^2 + D where D = -(b^2 - 4*a*c)/(4*a)
    • x + b/(2*a) = u and du = dx
    • ∫(1/(a*x^2 + b*x + c))dx
    • ∫ = (1/(D + a*u^2)du
    • = (1/Sqr(a*D)*arctan(Sqr(D/a)*u)
    • = (1/Sqr(a*D)*arctan(Sqr(D/a)*(x + b/(2*a))
    • Hence A = 1/Sqr(a*D), B = Sqr(D/a) and C = b/(2*a)
Example : Find ∫(1/(x^2 + 4*x + 5))dx
  • Since x^2 + 4*x + 5 = (x + 2)^2 + 1
  • Hence ∫(1/(x^2 + 4*x + 4))dx ∫(1/(1 + (x + 2)^2)dx
  • = arctan(x + 2)

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CA 16 06. ∫(1/(a*x^2 + b*x + c))dx = A*ln(x+B) + C*ln(x+D) : case 3

Case 2 : b^2 - 4*a*c GT 0
  • ∫(1/(a*x^2 + b*x + c))dx = A*ln(x+B) + C*ln(x+D)
  • Find A, B, C and D
    • Using completing the square
    • (a*x^2 + b*x + c) = a*(x^2 + b*x/a + (b/(2*a))^2 - (b/(2*a))^2 + c/a)
    • = a*(x + b/(2*a))^2 - (b^2 - 4*a*c)/(4*a)
    • Since b^2 - 4*a*c LT 0
    • Hence (a*x^2 + b*x + c)= a*(x + b/(2*a))^2 - D where D = (b^2 - 4*a*c)/(4*a)
    • x + b/(2*a) = u and du = dx
    • ∫(1/(a*x^2 + b*x + c))dx
    • ∫ =(1/a)*(1/(D/a - u^2)du
    • ∫ = (1/a)*(E/(Sqr(D/a) + u) + F/(Sqr(D/a) - u))du
    • = (E*ln(Sqr(D/a) + u) + F*ln(Sqr(D/a) - u))/a
    • = (E/a)*ln(Sqr(D/a) + (x+b/(2*a))) + (F/a)*ln(Sqr(D/a) - (x+b/(2*a)))
    • Hence A = E/a, B = Sqr(D/a) + b/(2*a), C = F/a and B = Sqr(D/a) - b/(2*a)
    • Where E and F can be found by partial fraction
Example : Find ∫(1/(x^2 - 3*x + 2))dx
  • Since x^2 - 3*x + 2 = (x - 1)*(x - 2)
  • Hence ∫(1/(x^2 + 4*x + 4))dx ∫(1/((x - 1)*(x - 2))dx
  • Using partial faction :
    • 1/((x-1)*(x-2)) = A/(x - 1) + B/(x - 2)
    • Multiply both sides by (x-1)*(x-2)
    • 1 = A*(x - 2) + B*(x - 1)
    • 1 = (A+B)*x - (2*A + B)
    • Hence A+B = 0 and 2*A + B = 1
    • Hence A = -B and 2*(-B) + B = 1. Hence B = -1 and A = 1
  • ∫(1/(x^2 - 3*x + 2))dx
  • = ∫(1/(x - 1) - 1/(x - 2))dx
  • = ln(x - 1) + ln(x - 2)

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CA 16 07. Answer

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CA 16 08. Answer

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CA 16 09. Answer


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CA 16 10. Answer


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CA 16 00. Outline

Integrals of ration function from anti-differentiation
  • ∫(1/(1 + x^2))dx = arctan(x)
  • ∫(1/(1 - x^2))dx = arctanh(x)
Integrals of ir-ration function from anti-differentiation
  • ∫(1/Sqr(1 - x^2))dx = arcsin(x)
  • ∫(1/Sqr(1 + x^2))dx = arcsinh(x)
  • ∫(1/(x*Sqr(x^2 - 1))dx = arcsec(x)
  • ∫(1/(x*Sqr(1 - x^2))dx = arcsech(x)
Integral formulae
  • 1. ∫(1/(a + b*x^2))dx = Sqr(1/a*b)*arctan(sqr(b/a)*x)
  • 2. ∫(1/(a + b*x^2))dx = Sqr(1/a*b)*arcsin(sqr(b/a)*x)
  • 3. ∫(1/(a*x^2 + b*x + c))dx = A/(x + B)
  • 4. ∫(1/(a*x^2 + b*x + c))dx = A*arctan(B*x)
  • 5. ∫(1/(a*x^2 + b*x + c))dx = A*ln(x-B)+B*ln(x-C)

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