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

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

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

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

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

Solution steps

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

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

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

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

Add a constant to the solution

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

\frac{d}{d x} (e^{x} (x - 3))

\int 8 (tan^{2} (x) + tan^{4} (x)) d x

x, \frac{1}{2} x, 9

x \to - 1 + lim (2 x^{3} - 2 m)

f (x) = 5 e^{2 x}

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