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

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

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integral of 5/(36+x^2)

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

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

Solution steps

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

\frac{\partial}{\partial y} (x y (1 - x - y))

y = x^{4} + 2 e^{x}

x \to 9 lim (∣ x + 3 ∣)

3 t \cdot \frac{d y}{d t} - y = t

\int x + e^{4 x} d x

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