What is the integral of (54)/(1+36x^2) ?

The integral of (54)/(1+36x^2) is 9arctan(6x)+C

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integral of (54)/(1+36x^2)

\int \frac{54}{1 + 36 x ^{2}} d x

9 arctan (6 x) + C

Solution steps

\int \frac{54}{1 + 36 x ^{2}} d x

Take the constant out:

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

= 54 \cdot \int \frac{1}{1 + 36 x ^{2}} d x

Apply Integral Substitution

= 54 \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

= 54 \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)

= 54 \cdot \frac{1}{6} arctan (u)

Substitute back

u = 6 x

= 54 \cdot \frac{1}{6} arctan (6 x)

Simplify

54 \cdot \frac{1}{6} arctan (6 x) : 9 arctan (6 x)

= 9 arctan (6 x)

Add a constant to the solution

= 9 arctan (6 x) + C

f (x) = arcsin (\frac{x}{2})

x \to 0 + lim (ln (x) + \frac{1}{x ^{2}})

\int sec (14 w) tan (14 w) d w

\int 2 (x - 2) ln (x - 2) d x

n = 0 \sum \infty \frac{1}{4 n ^{2} - 1}

What is the integral of (54)/(1+36x^2) ?
The integral of (54)/(1+36x^2) is 9arctan(6x)+C