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

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Answer 1 · 2015-09-28T00:34:46

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

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

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

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.

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