Question

How do you find the definite integral of `intx^3 sqrt(x^2 + 1 )dx` from `[0,sqrt3]`?

Answer 1

`58/15`

Explanation:

Substitute `x^2+1 = u^2`. Then `2xdx = 2udu` and the limits `x=0` and `x = sqrt(3)` becomes `u=1` and `u=2`, respectively.

Thus

`int_0^{sqrt(3)}x^3 sqrt(x^2 + 1 )dx = int_0^{sqrt(3)}x^2 sqrt(x^2 + 1 )quad xdx`
`qquad = quad int_1^2 (u^2-1)u times udu = int_1^2 u^4 du - int_1^2 u^2du`
`qquad = (u^5/5-u^3/3)_1^2 = (32/5-8/3)-(1/5-1/3)`
`qquad = 56/15-(2/15) = 58/15`