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

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

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

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

\frac{2 x ^{3}}{3} - 6 x + 63 arctan (\frac{x}{3}) + C

Solution steps

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

Take the constant out:

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

= 2 \cdot \int \frac{x ^{4}}{x ^{2} + 3} d x

Apply Integral Substitution

= 2 \cdot \int \frac{3 3 u ^{4}}{u ^{2} + 1} d u

Take the constant out:

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

= 2 \cdot 33 \cdot \int \frac{u ^{4}}{u ^{2} + 1} d u

Long division

\frac{u ^{4}}{u ^{2} + 1} : u^{2} - 1 + \frac{1}{u ^{2} + 1}

= 2 \cdot 33 \cdot \int u^{2} - 1 + \frac{1}{u ^{2} + 1} d u

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= 2 \cdot 33 (\int u^{2} d u - \int 1 d u + \int \frac{1}{u ^{2} + 1} d u)

\int u^{2} d u = \frac{u ^{3}}{3}

\int 1 d u = u

\int \frac{1}{u ^{2} + 1} d u = arctan (u)

= 2 \cdot 33 (\frac{u ^{3}}{3} - u + arctan (u))

Substitute back

u = \frac{x}{3}

= 2 \cdot 33 \frac{( \frac{x}{3} ) ^{3}}{3} - \frac{x}{3} + arctan (\frac{x}{3})

Simplify

2 \cdot 33 \frac{( \frac{x}{3} ) ^{3}}{3} - \frac{x}{3} + arctan (\frac{x}{3}) : \frac{2 x ^{3}}{3} - 6 x + 63 arctan (\frac{x}{3})

= \frac{2 x ^{3}}{3} - 6 x + 63 arctan (\frac{x}{3})

Add a constant to the solution

= \frac{2 x ^{3}}{3} - 6 x + 63 arctan (\frac{x}{3}) + C

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