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

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

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

integral

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

- \frac{1}{x} + ln ∣ x + 1 ∣ + C

Solution steps

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

Expand

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

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

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

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

= - ln ∣ x ∣ - \frac{1}{x} + ln ∣ x + 1 ∣ + ln ∣ x ∣ - ln ∣ x + 1 ∣ + ln ∣ 1 + x ∣

Simplify

= - \frac{1}{x} + ln ∣ x + 1 ∣

Add a constant to the solution

= - \frac{1}{x} + ln ∣ x + 1 ∣ + C

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