What is the integral of (x^3)/(2x+1) ?

The integral of (x^3)/(2x+1) is (x^3)/6-(x^2)/8+1/8 x-1/16 ln|2x+1|+C

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integral of (x^3)/(2x+1)

\int \frac{x ^{3}}{2 x + 1} d x

\frac{x ^{3}}{6} - \frac{x ^{2}}{8} + \frac{1}{8} x - \frac{1}{16} ln ∣ 2 x + 1 ∣ + C

Solution steps

\int \frac{x ^{3}}{2 x + 1} d x

Long division

\frac{x ^{3}}{2 x + 1} : \frac{x ^{2}}{2} - \frac{x}{4} + \frac{1}{8} - \frac{1}{8 ( 2 x + 1 )}

= \int \frac{x ^{2}}{2} - \frac{x}{4} + \frac{1}{8} - \frac{1}{8 ( 2 x + 1 )} 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

= \int \frac{x ^{2}}{2} d x - \int \frac{x}{4} d x + \int \frac{1}{8} d x - \int \frac{1}{8 ( 2 x + 1 )} d x

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

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

\int \frac{1}{8} d x = \frac{1}{8} x

\int \frac{1}{8 ( 2 x + 1 )} d x = \frac{1}{16} ln ∣ 2 x + 1 ∣

= \frac{x ^{3}}{6} - \frac{x ^{2}}{8} + \frac{1}{8} x - \frac{1}{16} ln ∣ 2 x + 1 ∣

Add a constant to the solution

= \frac{x ^{3}}{6} - \frac{x ^{2}}{8} + \frac{1}{8} x - \frac{1}{16} ln ∣ 2 x + 1 ∣ + C

\frac{d}{d x} (ln (a x^{2} + b))

y \to 1 lim (85)

\frac{d}{d y} (2 e^{x y^{2}})

y^{^{'}} = 3 y + x

f (x) = \frac{x}{1 + x}

What is the integral of (x^3)/(2x+1) ?
The integral of (x^3)/(2x+1) is (x^3)/6-(x^2)/8+1/8 x-1/16 ln|2x+1|+C