What is the integral of 4/(1+9x^2) ?

The integral of 4/(1+9x^2) is 4/3 arctan(3x)+C

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integral of 4/(1+9x^2)

\int \frac{4}{1 + 9 x ^{2}} d x

\frac{4}{3} arctan (3 x) + C

Solution steps

\int \frac{4}{1 + 9 x ^{2}} d x

Take the constant out:

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

= 4 \cdot \int \frac{1}{1 + 9 x ^{2}} d x

Apply Integral Substitution

= 4 \cdot \int \frac{1}{3 ( 1 + u ^{2} )} d u

Take the constant out:

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

= 4 \cdot \frac{1}{3} \cdot \int \frac{1}{1 + u ^{2}} d u

Use the common integral:

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

= 4 \cdot \frac{1}{3} arctan (u)

Substitute back

u = 3 x

= 4 \cdot \frac{1}{3} arctan (3 x)

Simplify

4 \cdot \frac{1}{3} arctan (3 x) : \frac{4}{3} arctan (3 x)

= \frac{4}{3} arctan (3 x)

Add a constant to the solution

= \frac{4}{3} arctan (3 x) + C

\int - x^{5} d x

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

\frac{d}{d x} (e^{s i n (x + y)})

\frac{d}{d x} (5 x - \frac{8}{x})

x \to - 3 lim (4 - x^{2})

What is the integral of 4/(1+9x^2) ?
The integral of 4/(1+9x^2) is 4/3 arctan(3x)+C