Question

How do you find the antiderivative of `int x(x^2+1)^100 dx`?

Answer 1

`(1/202)(x^2+1)^101+C`

Explanation:

Write down `(x^2+1)^101+C` as a guess and differentiate it (chain rule). You get `(101).(x^2+1)^(101-1).(2x)=202(x^2+1)^100`. This is close, but too big by a factor of `202`. so divide the first guess by `202`.

Alternatively, substitute `u=x^2+1`, giving `(du)/(dx)=2x`, `dx/(du)=1/(2x)`.

Then by the substitution formula the integral becomes
`int cancel(x).u^100.(1/(2 cancel(x)))du`
`=(1/2)intu^100du`
`=(1/2)(1/(100+1))u^(100+1)+C` by the power law
`=(1/202)(x^2+1)^101+C`

Whatever you do, don't try to expand the bracket by the binomial expansion!