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

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

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

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

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

Solution steps

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

Apply u-substitution

: \int \frac{2}{u ^{2} + 1} d u

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

Take the constant out:

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

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

Use the common integral:

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

= 2 arctan (u)

Substitute back

u = \frac{x}{2}

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

Add a constant to the solution

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

\frac{d}{d x} (\frac{e ^{2 x} - 1}{x})

\frac{d}{d x} (a y (x))

\frac{d}{d x} (\frac{6}{x ^{5}})

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

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