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

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

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

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

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

Solution steps

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

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

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

Use the common integral:

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

= \frac{1}{2} arctan (x)

Add a constant to the solution

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

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

(5 x - 3) y^{^{''}} + y^{^{'}} = 0

6

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

x \to - 6 + lim (\frac{- 9 x}{x + 6})

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