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

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

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

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

\frac{1}{2} ln x^{2} + 1 + x + C

Solution steps

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

\frac{x ^{2} + x + 1}{x ^{2} + 1} = \frac{x}{x ^{2} + 1} + 1

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

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

\int 1 d x = x

= \frac{1}{2} ln x^{2} + 1 + x

Add a constant to the solution

= \frac{1}{2} ln x^{2} + 1 + x + C

\frac{3000 - 10 x ^{2} + 1}{3 x ^{3}}

\frac{1}{2} sin (2 t)

\frac{d}{d x} (\frac{1 + x}{1 - x})

7 x - 3

x \to 0 lim (\frac{e ^{x} - 1}{8 x})

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