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

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

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

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

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

Solution steps

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

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

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

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

\int 1 d x = x

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

Add a constant to the solution

= \frac{1}{2} ln x^{2} + 1 - arctan (x) + x + C

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