What is the integral of (3x^2-5x+1)/(2x+1) ?

The integral of (3x^2-5x+1)/(2x+1) is 3/4 x^2-13/4 x+17/8 ln|2x+1|-5/4-9/16+C

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integral of (3x^2-5x+1)/(2x+1)

\int \frac{3 x ^{2} - 5 x + 1}{2 x + 1} d x

\frac{3}{4} x^{2} - \frac{13}{4} x + \frac{17}{8} ln ∣ 2 x + 1 ∣ - \frac{5}{4} - \frac{9}{16} + C

Solution steps

\int \frac{3 x ^{2} - 5 x + 1}{2 x + 1} d x

Expand

\frac{3 x ^{2} - 5 x + 1}{2 x + 1} : \frac{3 x ^{2}}{2 x + 1} - \frac{5 x}{2 x + 1} + \frac{1}{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{3 x ^{2}}{2 x + 1} d x - \int \frac{5 x}{2 x + 1} d x + \int \frac{1}{2 x + 1} d x

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

\int \frac{5 x}{2 x + 1} d x = \frac{5}{4} (2 x + 1 - ln ∣ 2 x + 1 ∣)

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

= \frac{3}{16} (4 x^{2} - 4 x + 2 ln ∣ 2 x + 1 ∣ - 3) - \frac{5}{4} (2 x + 1 - ln ∣ 2 x + 1 ∣) + \frac{1}{2} ln ∣ 2 x + 1 ∣

Simplify

\frac{3}{16} (4 x^{2} - 4 x + 2 ln ∣ 2 x + 1 ∣ - 3) - \frac{5}{4} (2 x + 1 - ln ∣ 2 x + 1 ∣) + \frac{1}{2} ln ∣ 2 x + 1 ∣ : \frac{3}{4} x^{2} - \frac{13}{4} x + \frac{17}{8} ln ∣ 2 x + 1 ∣ - \frac{5}{4} - \frac{9}{16}

= \frac{3}{4} x^{2} - \frac{13}{4} x + \frac{17}{8} ln ∣ 2 x + 1 ∣ - \frac{5}{4} - \frac{9}{16}

Add a constant to the solution

= \frac{3}{4} x^{2} - \frac{13}{4} x + \frac{17}{8} ln ∣ 2 x + 1 ∣ - \frac{5}{4} - \frac{9}{16} + C

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What is the integral of (3x^2-5x+1)/(2x+1) ?
The integral of (3x^2-5x+1)/(2x+1) is 3/4 x^2-13/4 x+17/8 ln|2x+1|-5/4-9/16+C