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

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

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

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

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

Solution steps

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

Simplify

= \int \frac{1}{4 + x ^{2}} d x

Apply Integral Substitution

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

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}{u ^{2} + 1} d u

Use the common integral:

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

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

Substitute back

u = \frac{x}{2}

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

Add a constant to the solution

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

\int \frac{x ^{2} + 1}{3 + 2 x - x ^{2}} d x

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\int \frac{x ^{2}}{x ^{2} + 4 x - 5} d x

\int x (1 - 2 x)^{4} d x

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