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

Question

4092 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-12-21T13:27:16

$int (3-x)7^((3-x)^2)dx=-(7^((3-x)^2))/(2ln7)+C$

Substitute $t=(3-x)^2$ , $dt=-2(3-x)dx$ , and consider that $7^alpha= e^(alphaln7)$ :

$int (3-x)7^((3-x)^2)dx= -1/2int e^(ln7t)dt=-1/2ln7e^(ln7t)+C=-(7^((3-x)^2))/(2ln7)+C$

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