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

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

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

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

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

Solution steps

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

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

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

Use the common integral:

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

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

Add a constant to the solution

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

1 - \frac{1}{t}

\int \frac{8}{8 + e ^{x}} d x

y^{2} = 9 x, at x = 16

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

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