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

The integral of (x^2-3x)/(2x+1) is 1/4 x^2-7/4 x+7/8 ln|2x+1|-3/4-3/16+C

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

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

\frac{1}{4} x^{2} - \frac{7}{4} x + \frac{7}{8} ln ∣ 2 x + 1 ∣ - \frac{3}{4} - \frac{3}{16} + C

Solution steps

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

Expand

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

\int \frac{x ^{2}}{2 x + 1} d x = \frac{1}{16} (4 x^{2} - 4 x + 2 ln ∣ 2 x + 1 ∣ - 3)

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

= \frac{1}{16} (4 x^{2} - 4 x + 2 ln ∣ 2 x + 1 ∣ - 3) - \frac{3}{4} (2 x + 1 - ln ∣ 2 x + 1 ∣)

Simplify

\frac{1}{16} (4 x^{2} - 4 x + 2 ln ∣ 2 x + 1 ∣ - 3) - \frac{3}{4} (2 x + 1 - ln ∣ 2 x + 1 ∣) : \frac{1}{4} x^{2} - \frac{7}{4} x + \frac{7}{8} ln ∣ 2 x + 1 ∣ - \frac{3}{4} - \frac{3}{16}

= \frac{1}{4} x^{2} - \frac{7}{4} x + \frac{7}{8} ln ∣ 2 x + 1 ∣ - \frac{3}{4} - \frac{3}{16}

Add a constant to the solution

= \frac{1}{4} x^{2} - \frac{7}{4} x + \frac{7}{8} ln ∣ 2 x + 1 ∣ - \frac{3}{4} - \frac{3}{16} + C

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What is the integral of (x^2-3x)/(2x+1) ?
The integral of (x^2-3x)/(2x+1) is 1/4 x^2-7/4 x+7/8 ln|2x+1|-3/4-3/16+C