What is the antiderivative of #pi^(x-1)dx#? Calculus Introduction to Integration Integrals of Exponential Functions 1 Answer Massimiliano Feb 19, 2015 The answer is: #pi^(x-1)/lnpi+c#. Since: #inta^f(x)*f'(x)dx=a^f(x)/lna+c# Than: #intpi^(x-1)dx=pi^(x-1)/lnpi+c#. Answer link Related questions How do you evaluate the integral #inte^(4x) dx#? How do you evaluate the integral #inte^(-x) dx#? How do you evaluate the integral #int3^(x) dx#? How do you evaluate the integral #int3e^(x)-5e^(2x) dx#? How do you evaluate the integral #int10^(-x) dx#? What is the integral of #e^(x^3)#? What is the integral of #e^(0.5x)#? What is the integral of #e^(2x)#? What is the integral of #e^(7x)#? What is the integral of #2e^(2x)#? See all questions in Integrals of Exponential Functions Impact of this question 5786 views around the world You can reuse this answer Creative Commons License