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

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

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

integral

\frac{- ( x ^{2} + x )}{2 x + 1}

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

Solution steps

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

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

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

Expand

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

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

Simplify

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

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

Add a constant to the solution

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

\frac{( 1 - x ^{8} )}{( 1 - x )}

f (8 - 2 x)

\frac{1}{1 + e ^{x}}

\frac{e ^{x}}{( e ^{x} + 2 )}

2 x (4 - x^{2})

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