Question

How do you find the indefinite integral of `int x^3(x^4+3)^2dx`?

Answer 1

`(x^4 +3)^2 /12 +C`

Explanation:

Let `x^4 +3 =u`, so that `x^3dx= 1/4 du` On substituting, the given integral becomes `int 1/4 u^2 du`

=`u^3/12 +c`

=`(x^4 +3)^2 /12 +C`

Answer 2

The answer is `x^12/12+3x^8/4+9x^4/4+c`.

Explanation:

First of all, expand the square: using the formula

`(a+b)^2= a^2 +2ab + b^2`

we have that

`(x^4+3)^2 = x^8 + 6x^4 + 9`

Now we can multiply the expanded square for `x^3`, obtaining

`x^3 * (x^8 + 6x^4 + 9) = x^11 + 6x^7 + 9x^3`

Now this quantity is easy to integrate, because of the linearity of the integral, which means that the integral of the sum is the sum of the integrals:

`\int x^11 + 6x^9 + 9x^3 dx = \int x^11 dx + \int 6x^7 + \int 9x^3 dx`

and each of these integrals can be done using the same rule, i.e.

`\int ax^n dx = {ax^{n+1}}/{n+1}`.

So: `\int x^11= x^12/12`,
`\int 6x^7=6x^8/8= 3x^8/4`, and
`\int 9x^3 = 9x^4/4`.

The answer is thus `x^12/12+3x^8/4+9x^4/4+c`.