What is the integral of 5/(x^2+36) ?

The integral of 5/(x^2+36) is 5/6 arctan(x/6)+C

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integral of 5/(x^2+36)

\int \frac{5}{x ^{2} + 36} d x

\frac{5}{6} arctan (\frac{x}{6}) + C

Solution steps

\int \frac{5}{x ^{2} + 36} d x

Take the constant out:

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

= 5 \cdot \int \frac{1}{x ^{2} + 36} d x

Apply Integral Substitution

= 5 \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

= 5 \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)

= 5 \cdot \frac{1}{6} arctan (u)

Substitute back

u = \frac{x}{6}

= 5 \cdot \frac{1}{6} arctan (\frac{x}{6})

Simplify

5 \cdot \frac{1}{6} arctan (\frac{x}{6}) : \frac{5}{6} arctan (\frac{x}{6})

= \frac{5}{6} arctan (\frac{x}{6})

Add a constant to the solution

= \frac{5}{6} arctan (\frac{x}{6}) + C

\int 2^{- x} d x

\int_{1}^{5} 17 d x

y^{^{'}} = e^{- y}

\frac{d}{d x} (arctan (6 x))

\int_{0}^{π} (8 e^{x} + 5 sin (x)) d x

What is the integral of 5/(x^2+36) ?
The integral of 5/(x^2+36) is 5/6 arctan(x/6)+C