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

The integral of (4x^3)/(2x+3) is 4((x^3)/6-(3x^2)/8+9/8 x-27/16 ln|2x+3|)+C

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

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

4 (\frac{x ^{3}}{6} - \frac{3 x ^{2}}{8} + \frac{9}{8} x - \frac{27}{16} ln ∣ 2 x + 3 ∣) + C

Solution steps

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

Take the constant out:

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

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

Long division

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

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

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

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

\int \frac{3 x}{4} d x = \frac{3 x ^{2}}{8}

\int \frac{9}{8} d x = \frac{9}{8} x

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

= 4 (\frac{x ^{3}}{6} - \frac{3 x ^{2}}{8} + \frac{9}{8} x - \frac{27}{16} ln ∣ 2 x + 3 ∣)

Add a constant to the solution

= 4 (\frac{x ^{3}}{6} - \frac{3 x ^{2}}{8} + \frac{9}{8} x - \frac{27}{16} ln ∣ 2 x + 3 ∣) + C

π r^{2} h

\int \frac{1}{x ^{2} x ^{2} - 9} d x

\frac{\partial}{\partial x} (4 x - 2 y + 3)

n = 1 \sum \infty \frac{x ^{n}}{n + 2}

x (\frac{d y}{d x}) = 7 x e^{x} - y + 9 x^{2}

What is the integral of (4x^3)/(2x+3) ?
The integral of (4x^3)/(2x+3) is 4((x^3)/6-(3x^2)/8+9/8 x-27/16 ln|2x+3|)+C