Question

How do i integrate (9x²+6x+6)(x³+x²+2x)?

Answer 1

`int(9x^2+6x+6)(x^3+x^2+2x)dx=3/2(x^3+x^2+2x)^2+C`

Explanation:

You could multiply everything out and apply the power rules for antidifferentiation, but that would be absurdly messy and tedious. Instead, we can perform a simple substitution:

`int(9x^2+6x+6)(x^3+x^2+2x)dx`

We'll let `u=x^3+x^2+2x`

Then,

`du=(3x^2+2x+2)dx`

`du` almost shows up in the integral, but not quite. If we multiply both sides by `3,` however, we get

`3du=3(3x^2+2x+2)dx`

`3du=(9x^2+6x+6)dx`

Which shows up in the integral! So this substitution is valid.

Rewrite the integral in terms of `u:`

`3intudu=3(1/2)u^2+C=3/2u^2+C`

Rewriting in terms of `x` yields

`int(9x^2+6x+6)(x^3+x^2+2x)dx=3/2(x^3+x^2+2x)^2+C`