Question

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

Answer 1

`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`