What is the integral of (8x^3+2x+2)/(2x+1) ?

The integral of (8x^3+2x+2)/(2x+1) is (4x^3)/3-x^2+2x+1/2+C

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integral of (8x^3+2x+2)/(2x+1)

\int \frac{8 x ^{3} + 2 x + 2}{2 x + 1} d x

\frac{4 x ^{3}}{3} - x^{2} + 2 x + \frac{1}{2} + C

Solution steps

\int \frac{8 x ^{3} + 2 x + 2}{2 x + 1} d x

Expand

\frac{8 x ^{3} + 2 x + 2}{2 x + 1} : \frac{8 x ^{3}}{2 x + 1} + \frac{2 x}{2 x + 1} + \frac{2}{2 x + 1}

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

\int \frac{8 x ^{3}}{2 x + 1} d x = \frac{4 x ^{3}}{3} - x^{2} + x - \frac{1}{2} ln ∣ 2 x + 1 ∣

\int \frac{2 x}{2 x + 1} d x = x + \frac{1}{2} - \frac{1}{2} ln ∣ 2 x + 1 ∣

\int \frac{2}{2 x + 1} d x = ln ∣ 2 x + 1 ∣

= \frac{4 x ^{3}}{3} - x^{2} + x - \frac{1}{2} ln ∣ 2 x + 1 ∣ + x + \frac{1}{2} - \frac{1}{2} ln ∣ 2 x + 1 ∣ + ln ∣ 2 x + 1 ∣

Simplify

\frac{4 x ^{3}}{3} - x^{2} + x - \frac{1}{2} ln ∣ 2 x + 1 ∣ + x + \frac{1}{2} - \frac{1}{2} ln ∣ 2 x + 1 ∣ + ln ∣ 2 x + 1 ∣ : \frac{4 x ^{3}}{3} - x^{2} + 2 x + \frac{1}{2}

= \frac{4 x ^{3}}{3} - x^{2} + 2 x + \frac{1}{2}

Add a constant to the solution

= \frac{4 x ^{3}}{3} - x^{2} + 2 x + \frac{1}{2} + C

\frac{d}{d x} (ln (\frac{x ^{2} sin ( x )}{2 x + 1}))

2 x^{3} + x^{2} - x - 1, x^{3} + 2 x^{2} + 5 x - 1

(2 x - 2)^{4} (x^{2} + x + 1)^{5}

\frac{x + 1}{x - 2}

\frac{d}{d x} (e^{\frac{1}{x ^{2}}})

What is the integral of (8x^3+2x+2)/(2x+1) ?
The integral of (8x^3+2x+2)/(2x+1) is (4x^3)/3-x^2+2x+1/2+C