What is the integral of 8/(36+x^2) ?

The integral of 8/(36+x^2) is 4/3 arctan(x/6)+C

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integral of 8/(36+x^2)

\int \frac{8}{36 + x ^{2}} d x

\frac{4}{3} arctan (\frac{x}{6}) + C

Solution steps

\int \frac{8}{36 + x ^{2}} d x

Take the constant out:

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

= 8 \cdot \int \frac{1}{36 + x ^{2}} d x

Apply Integral Substitution

= 8 \cdot \int \frac{1}{6 ( u ^{2} + 1 )} d u

Take the constant out:

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

= 8 \cdot \frac{1}{6} \cdot \int \frac{1}{u ^{2} + 1} d u

Use the common integral:

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

= 8 \cdot \frac{1}{6} arctan (u)

Substitute back

u = \frac{x}{6}

= 8 \cdot \frac{1}{6} arctan (\frac{x}{6})

Simplify

8 \cdot \frac{1}{6} arctan (\frac{x}{6}) : \frac{4}{3} arctan (\frac{x}{6})

= \frac{4}{3} arctan (\frac{x}{6})

Add a constant to the solution

= \frac{4}{3} arctan (\frac{x}{6}) + C

f (x) = 5 - x

\frac{\partial}{\partial x} (\frac{6 x}{3 x ^{2} + 2 y ^{2} + 9})

\int_{1}^{\infty} 4 e^{- 2 x} d x

\frac{d}{d x} (3 x^{2} + 4 x)

\frac{d}{d x} ((x - 1) \cdot e^{x^{2} - 4})

What is the integral of 8/(36+x^2) ?
The integral of 8/(36+x^2) is 4/3 arctan(x/6)+C