What is the integral of (x^2+4)^{-1} ?

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

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integral of (x^2+4)^{-1}

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

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

Solution steps

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

n = 1 \sum \infty \frac{n - 1}{n 4 ^{n}}

\int \frac{24}{1 - cos ( 6 x )} d x

sin (x) + cos (x)

\int_{0}^{4} (1 + 2 x)^{2} d x

(1, 2), (3, 16)

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