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

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

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

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

- \frac{1}{3} arctan (3 x) + C

Solution steps

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

Take the constant out:

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

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

Apply Integral Substitution

= - \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

= - \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)

= - \frac{1}{3} arctan (u)

Substitute back

u = 3 x

= - \frac{1}{3} arctan (3 x)

Add a constant to the solution

= - \frac{1}{3} arctan (3 x) + C

\int (2 x + 2)^{- 1} d x

\int - 5 x^{2} sin (π x) d x

\frac{1}{( x ^{2} - 1 ) ( x ^{2} + 5 x + 5 )}

\int_{0}^{1} (e^{2 x} - e^{- x}) d x

x \to 1 lim (\frac{3}{1 - x ^{3}})

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