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

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

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

integral

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

1 - x - 2 ln ∣ 1 - x ∣ + ln ∣ x ∣ - ln ∣ x - 1 ∣ + C

Solution steps

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

Expand

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

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

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

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

= 1 - x - ln ∣ 1 - x ∣ - ln ∣ 1 - x ∣ + ln ∣ x ∣ - ln ∣ x - 1 ∣

Simplify

= 1 - x - 2 ln ∣ 1 - x ∣ + ln ∣ x ∣ - ln ∣ x - 1 ∣

Add a constant to the solution

= 1 - x - 2 ln ∣ 1 - x ∣ + ln ∣ x ∣ - ln ∣ x - 1 ∣ + C

sin (2 x + 3) + cos (3 x - 2)

sin^{2} (5 x)

1 - sin^{2} (x)

1 - ∣ x ∣

(1 - x)^{3}

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