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

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

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

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

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

Solution steps

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

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

= \int \frac{1}{2} \cdot \frac{1}{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

- t

\frac{\partial}{\partial x} (- 2 s t)

x \to 0 + lim (\frac{x}{cos ( 4 x ) - 1})

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

\frac{1}{x + 2}

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