What is the integral of (3x^4-12)/(x^2+2) ?

The integral of (3x^4-12)/(x^2+2) is 3((x^3)/2-(x^3)/6-2x)+C

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integral of (3x^4-12)/(x^2+2)

\int \frac{3 x ^{4} - 12}{x ^{2} + 2} d x

3 (\frac{x ^{3}}{2} - \frac{x ^{3}}{6} - 2 x) + C

Solution steps

\int \frac{3 x ^{4} - 12}{x ^{2} + 2} d x

Factor

\frac{3 x ^{4} - 12}{x ^{2} + 2} : 3 (x + 2) (x - 2)

= \int 3 (x + 2) (x - 2) d x

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= 3 \cdot \int (x + 2) (x - 2) d x

Apply Integration By Parts

: (x + 2) (\frac{x ^{2}}{2} - 2 x) - \int \frac{x ^{2}}{2} - 2 x d x

= 3 ((x + 2) (\frac{x ^{2}}{2} - 2 x) - \int \frac{x ^{2}}{2} - 2 x d x)

\int \frac{x ^{2}}{2} - 2 x d x = \frac{x ^{3}}{6} - \frac{x ^{2}}{2}

= 3 ((x + 2) (\frac{x ^{2}}{2} - 2 x) - (\frac{x ^{3}}{6} - \frac{x ^{2}}{2}))

Expand

(x + 2) (\frac{x ^{2}}{2} - 2 x) - (\frac{x ^{3}}{6} - \frac{x ^{2}}{2}) : \frac{x ^{3}}{2} - \frac{x ^{3}}{6} - 2 x

= 3 (\frac{x ^{3}}{2} - \frac{x ^{3}}{6} - 2 x)

Add a constant to the solution

= 3 (\frac{x ^{3}}{2} - \frac{x ^{3}}{6} - 2 x) + C

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What is the integral of (3x^4-12)/(x^2+2) ?
The integral of (3x^4-12)/(x^2+2) is 3((x^3)/2-(x^3)/6-2x)+C