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How do you integrate #int (3-x)7^((3-x) ^2)dx#?

1 Answer

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#

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