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

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

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

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

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

Solution steps

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

Expand

\frac{x ^{2} - a x}{2 x + 1} : \frac{x ^{2}}{2 x + 1} - \frac{a 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{a 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{a x}{2 x + 1} d x = a \frac{1}{4} (2 x + 1 - ln ∣ 2 x + 1 ∣)

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

Add a constant to the solution

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

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