What is the integral of (2x^5)/(3x^2+4) ?

The integral of (2x^5)/(3x^2+4) is 2((x^4}{12}-\frac{2x^2)/9+8/27 ln|3x^2+4|)+C

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integral of (2x^5)/(3x^2+4)

\int \frac{2 x ^{5}}{3 x ^{2} + 4} d x

2 (\frac{x ^{4}}{12} - \frac{2 x ^{2}}{9} + \frac{8}{27} ln 3 x^{2} + 4) + C

Solution steps

\int \frac{2 x ^{5}}{3 x ^{2} + 4} 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 ^{5}}{3 x ^{2} + 4} d x

Long division

\frac{x ^{5}}{3 x ^{2} + 4} : \frac{x ^{3}}{3} - \frac{4 x}{9} + \frac{16 x}{9 ( 3 x ^{2} + 4 )}

= 2 \cdot \int \frac{x ^{3}}{3} - \frac{4 x}{9} + \frac{16 x}{9 ( 3 x ^{2} + 4 )} d x

Apply the Sum Rule:

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

= 2 (\int \frac{x ^{3}}{3} d x - \int \frac{4 x}{9} d x + \int \frac{16 x}{9 ( 3 x ^{2} + 4 )} d x)

\int \frac{x ^{3}}{3} d x = \frac{x ^{4}}{12}

\int \frac{4 x}{9} d x = \frac{2 x ^{2}}{9}

\int \frac{16 x}{9 ( 3 x ^{2} + 4 )} d x = \frac{8}{27} ln 3 x^{2} + 4

= 2 (\frac{x ^{4}}{12} - \frac{2 x ^{2}}{9} + \frac{8}{27} ln 3 x^{2} + 4)

Add a constant to the solution

= 2 (\frac{x ^{4}}{12} - \frac{2 x ^{2}}{9} + \frac{8}{27} ln 3 x^{2} + 4) + C

\frac{d}{d x} (sin (- 1 x))

\int 3 x^{2} + 6 x d x

\frac{d}{d x} (100 e^{- 2 x})

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

\frac{d y}{d x} = 0.276 y - 0.0012 y^{2}

What is the integral of (2x^5)/(3x^2+4) ?
The integral of (2x^5)/(3x^2+4) is 2((x^4}{12}-\frac{2x^2)/9+8/27 ln|3x^2+4|)+C