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

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

\frac{- 3 e ^{(- s 2)}}{( s ^{2} - 5 s + 6 )}

x \to 0 + lim (x^{3} \cdot ln (x))

ln (x + 8 + x^{2})

\frac{1}{2 x ^{2}}

(1 - t^{2}) y^{^{''}} + 2 t y^{^{'}} - 2 y = 0

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