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

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

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

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

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

Solution steps

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

Expand

\frac{1 - x ^{2}}{1 - 2 x} : \frac{1}{1 - 2 x} - \frac{x ^{2}}{1 - 2 x}

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

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

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

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

Simplify

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

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

Add a constant to the solution

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

\int_{\frac{1}{2}}^{3} 16 x ln (2 x) d x

\frac{\partial}{\partial x} (ln (\frac{r}{ro}))

\frac{d}{d x} (e^{j x})

y^{^{'}} = y^{2} - 9

\frac{d}{d x} (x^{x^{x}})

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