Question

How do you find the indefinite integral of `int 1/4(x)(7 + 6x^2)dx`?

Answer 1

I would factor out the `1/4` then distribute `x` and integrate term by term.

Explanation:

`int 1/4(x)(7 + 6x^2)dx = 1/4 int (7x+6x^3)dx`

`= 1/4[(7x^2)/2+(6x^4)/4] +C`

The answer may be rewritten to taste.

Method 2 Do not factor out, but distribute the `1/4`

`int 1/4(x)(7 + 6x^2)dx = int ((7x)/4+(6x^3)/4)dx`

`= (7x^2)/8 + (6x^4)/16 +C`

Method 3 Use substitution.

`int 1/4(x)(7 + 6x^2)dx = 1/4 int (7+6x^2) x dx`

Let `u = 7+6x^2`, so that `du = 12x dx` and `xdx = (du)/12`

The integral becomes:

`1/4 int u (du)/12 = 1/48 int u du`

`= 1/48 u^2/2 +C`

`= 1/96(7+6x^2)^2 +C`

This method has an additional constant of `49/96` in it that is "absorbed" into the arbitrary constant `C` in the other solutions.