integral of x/(1-2x) — step-by-step solution

1) Let's integrate:

\int(\frac{x}{1 - 2 x})dx

Perform polynomial long division of

(x)

(- 2 x + 1)

Current expression	Divide leading terms	Quotient	Quotient × divisor	Remainder after subtracting
$x$	$\frac{x}{- 2 x}$	$- \frac{1}{2}$	$- \frac{1}{2} \cdot (- 2 x + 1) = x - \frac{1}{2}$	$x - (x - \frac{1}{2}) = \frac{1}{2}$

So the result of the euclidean division is:

- \frac{1}{2}

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

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

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

\frac{1}{2 \cdot (1 - 2 x)} - \frac{1}{2}

2) Let's compute the integral:

\int(\frac{1}{2 \cdot (1 - 2 x)} - \frac{1}{2})dx

\int(\frac{1}{2 \cdot (1 - 2 x)})dx - \int(\frac{1}{2})dx

Let

u = 2 - 4 x

. Then

du = -4 dx

, so

dx = \frac{du}{-4}

\int(\frac{1}{2 \cdot (1 - 2 x)})dx = - \frac{1}{4} \cdot \int(\frac{1}{u})du

We use

\int(\frac{1}{u})du = \ln | u |

. The absolute value covers both signs of the denominator, on any interval where

2 - 4 x \neq 0

Substitute

u

2 - 4 x

back into the antiderivative

ℹ

\int(\frac{1}{2 \cdot (1 - 2 x)})dx

- \frac{1}{4} \cdot \ln | 2 - 4 x |

‖

- \int(\frac{1}{2})dx

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

- \frac{1}{4} \cdot \ln | 2 - 4 x | - \frac{1}{2} \cdot x

ℹ

- \frac{1}{4} \cdot \ln | 2 - 4 x |

- \frac{\ln | 2 - 4 x |}{4}

‖

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

- \frac{x}{2}

‖

- \frac{\ln | 2 - 4 x |}{4} - \frac{x}{2}

\frac{- \ln | 2 - 4 x | + 2 \cdot (- x)}{4}

\frac{- \ln | 2 - 4 x | + 2 \cdot (- x)}{4}

- \frac{\ln | 2 - 4 x | + 2 x}{4}

▶Indefinite integrals are defined up to an additive constant C, so this yields:

- \frac{2 x + \ln | 2 - 4 x |}{4} + C