Question

How do you integrate `int (x+3)(x-1)^5` using substitution?

Answer 1

`int (x+3)(x-1)^5dx =1/7(x-1)^6(x+11/3)+C`

Explanation:

You could expand the power of the binomial and integrate using the power rule.

To avoid the burdensome calculation of a fifth power, substitute:

`t= (x-1)`
`dt = dx`

so that:

`int (x+3)(x-1)^5dx = int (t+4)t^5dt = int t^6dt + 4 intt^5dt = t^7/7+ 2/3t^6 +C`

Now substitute back `x`:

`int (x+3)(x-1)^5dx = 1/7(x-1)^7 +2/3(x-1)^6+C = 1/7(x-1)^6(x-1 + 14/3) + C = 1/7(x-1)^6(x+11/3)+C`