What is the integral of 1/((x^2+4)) ?

The integral of 1/((x^2+4)) is 1/2 arctan(x/2)+C

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integral of 1/((x^2+4))

integral

\frac{1}{( x ^{2} + 4 )}

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

Solution steps

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

\frac{1}{( a ^{2} - x ^{2} ) ^{\frac{1}{2}}}

\frac{e ^{1}}{x}

1 + tan^{2} (x)

\frac{1}{a ^{2} - x ^{2}}

\frac{- 1}{( x + 1 ) ^{2}}

What is the integral of 1/((x^2+4)) ?
The integral of 1/((x^2+4)) is 1/2 arctan(x/2)+C