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Integrals Involving Logarithmic Functions quiz

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이 집합의 용어 (15)

  • What is the integral of 1/x dx?
    The integral of 1/x dx is ln(|x|) + c, where c is the constant of integration.
  • Why do we use absolute value bars in the integral of 1/x?
    Absolute value bars ensure the antiderivative is defined for all x except zero, including negative values.
  • How can 1/x be rewritten to use the power rule for integration?
    1/x can be rewritten as x^(-1), but the power rule does not apply for n = -1; instead, it results in a logarithmic function.
  • What is the integral of 5/x dx?
    The integral of 5/x dx is 5 ln(|x|) + c.
  • How do you integrate 1/x^2 dx?
    Rewrite 1/x^2 as x^(-2) and use the power rule, resulting in -1/x + c.
  • What is the integral of 3/x dx?
    The integral of 3/x dx is 3 ln(|x|) + c.
  • How do you integrate a function like 4/(3+4x) dx?
    Use substitution by setting u = 3 + 4x, then the integral becomes ln(|3+4x|) + c.
  • What substitution should you use for integrating 2x+4 over x^2+4x dx?
    Set u = x^2 + 4x, so du = 2x + 4 dx, and the integral becomes ln(|x^2+4x|) + c.
  • What is the result of integrating 2x+4/(x^2+4x) dx?
    The result is ln(|x^2+4x|) + c.
  • How do you approach integrating 5 dx over x ln(x)^3?
    Set u = ln(x), so du = 1/x dx, and rewrite the integral as 5 ∫ u^(-3) du.
  • What is the integral of 5 dx over x ln(x)^3?
    The integral is -5/2 (ln(x))^(-2) + c.
  • What substitution is useful when integrating functions with an 'inside' function like ln(x)?
    Set u equal to the inside function, such as u = ln(x), to simplify the integral.
  • How do you integrate secant x dx using substitution?
    Multiply by (secant x + tangent x)/(secant x + tangent x), set u = secant x + tangent x, and the integral becomes ln(|secant x + tangent x|) + c.
  • What is the integral of cosecant x dx?
    The integral is -ln(|cosecant x + cotangent x|) + c.
  • Why might you multiply by a fraction like secant x + tangent x over itself when integrating secant x?
    Multiplying by this fraction is equivalent to multiplying by 1 and helps set up a substitution that simplifies the integral.