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

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

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

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

\frac{1}{6} arctan (6 x) + C

Solution steps

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

Apply Integral Substitution

= \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

= \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)

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

Substitute back

u = 6 x

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

Add a constant to the solution

= \frac{1}{6} arctan (6 x) + C

(tan (θ))^{^{'}}

x = y^{3}, x = y^{2}

\int \frac{( x + 2 )}{x ^{2} + 4 x + 5} d x

\frac{d}{d x} (- 4 x y)

\int tan^{3} (x) sin (x) d x

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