What is the integral of (36)/(1+36x^2) ?

The integral of (36)/(1+36x^2) is 6arctan(6x)+C

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integral of (36)/(1+36x^2)

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

6 arctan (6 x) + C

Solution steps

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

Take the constant out:

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

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

Apply Integral Substitution

= 36 \cdot \int \frac{1}{6 ( 1 + u ^{2} )} d u

Take the constant out:

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

= 36 \cdot \frac{1}{6} \cdot \int \frac{1}{1 + u ^{2}} d u

Use the common integral:

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

= 36 \cdot \frac{1}{6} arctan (u)

Substitute back

u = 6 x

= 36 \cdot \frac{1}{6} arctan (6 x)

Simplify

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

= 6 arctan (6 x)

Add a constant to the solution

= 6 arctan (6 x) + C

y^{^{'}} = 1 - 2 y

f (x) = \frac{x}{1 - ln ( x - 4 )}

f (x) = (1 + 3 x) (5 - 4 x), at x = 1

\frac{d}{d x} (a cos (a x))

\int \frac{x ^{3} + 4}{x ^{2} + 4} d x

What is the integral of (36)/(1+36x^2) ?
The integral of (36)/(1+36x^2) is 6arctan(6x)+C