5.6 Integrals Involving Exponential and Logarithmic Functions
3 min read•june 24, 2024
Exponential and logarithmic functions are key players in calculus. They pop up in many real-world scenarios, from population growth to compound interest. Mastering their integration techniques is crucial for solving complex problems.
These functions have unique properties that make them stand out. Their integrals often involve themselves or their inverses, creating elegant solutions. Understanding these patterns helps simplify seemingly tricky integrals and builds a strong foundation for advanced calculus concepts.
Integration Techniques for Exponential and Logarithmic Functions
Integrals of exponential functions
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Integrate the ex results in ex+C, where C is the constant of integration
Integrate exponential functions with bases other than e using the formula ∫axdx=lnaax+C, where a>0 and a=1 (a=2,10)
Multiply the result by the constant when integrating exponential functions multiplied by a constant k (k=3,−5)
∫k⋅exdx=k⋅ex+C
∫k⋅axdx=lnak⋅ax+C
Integrate exponential functions multiplied by a linear term x using integration by parts
∫x⋅exdx=(x−1)⋅ex+C
∫x⋅axdx=lnax⋅ax−(lna)2ax+C
Integration of logarithmic expressions
Integrate the natural logarithm lnx using the formula ∫lnxdx=xlnx−x+C
Multiply the result by the constant when integrating natural logarithms multiplied by a constant k (k=2,−3)
∫k⋅lnxdx=k⋅(xlnx−x)+C
Integrate logarithmic functions with bases other than e using the change of base formula
∫logaxdx=lnaxlnx−x+C, where a>0 and a=1 (a=2,10)
Integrate logarithmic functions multiplied by a linear term x using integration by parts
∫x⋅lnxdx=2x2lnx−4x2+C
∫x⋅logaxdx=lnax2lnx−2x2+C
Substitution for exponential and logarithmic integrals
Apply substitution to exponential functions by letting u equal the exponent and adjusting du and dx accordingly
Let u=2x, then du=2dx or dx=2du
Rewrite the integral in terms of u and simplify
Integrate with respect to u and substitute back the original variable
∫e2x⋅2dx=∫eu⋅2du=21∫eudu=21eu+C=21e2x+C
Apply substitution to logarithmic functions by letting u equal the argument of the logarithm and adjusting du and dx
Let u=3x, then du=3dx or dx=3du
Rewrite the integral in terms of u and simplify
Integrate with respect to u and substitute back the original variable