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

The integral of (3x^4-12)/(x^2+2)x is 1/12 (3x^2-6)^2+C

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

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

\frac{1}{12} (3 x^{2} - 6)^{2} + C

Solution steps

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

\frac{3 x ^{4} - 12}{x ^{2} + 2} = 3 x^{2} - 6

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

Apply u-substitution

: \int \frac{u}{6} d u

= \int \frac{u}{6} d u

Take the constant out:

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

= \frac{1}{6} \cdot \int u d u

Apply the Power Rule

: \frac{u ^{2}}{2}

= \frac{1}{6} \cdot \frac{u ^{2}}{2}

Substitute back

u = 3 x^{2} - 6

= \frac{1}{6} \cdot \frac{( 3 x ^{2} - 6 ) ^{2}}{2}

Simplify

\frac{1}{6} \cdot \frac{( 3 x ^{2} - 6 ) ^{2}}{2} : \frac{1}{12} (3 x^{2} - 6)^{2}

= \frac{1}{12} (3 x^{2} - 6)^{2}

Add a constant to the solution

= \frac{1}{12} (3 x^{2} - 6)^{2} + C

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What is the integral of (3x^4-12)/(x^2+2)x ?
The integral of (3x^4-12)/(x^2+2)x is 1/12 (3x^2-6)^2+C