How do you integrate #int (3-x)7^((3-x) ^2)dx#? Calculus Introduction to Integration Integrals of Exponential Functions 1 Answer Andrea S. Dec 21, 2016 #int (3-x)7^((3-x)^2)dx=-(7^((3-x)^2))/(2ln7)+C# Explanation: Substitute #t=(3-x)^2#, #dt=-2(3-x)dx#, and consider that #7^alpha= e^(alphaln7)#: #int (3-x)7^((3-x)^2)dx= -1/2int e^(ln7t)dt=-1/2ln7e^(ln7t)+C=-(7^((3-x)^2))/(2ln7)+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 3772 views around the world You can reuse this answer Creative Commons License