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

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

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

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

\frac{1}{3 2} arctan (\frac{3}{2} x) + C

Solution steps

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

Apply Integral Substitution

= \int \frac{1}{3 2 ( 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 2} \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 2} arctan (u)

Substitute back

u = \frac{3}{2} x

= \frac{1}{3 2} arctan (\frac{3}{2} x)

Add a constant to the solution

= \frac{1}{3 2} arctan (\frac{3}{2} x) + C

\frac{d y}{d x} = 9 x

\int \frac{y}{1 + y} d y

\int sec (3 x) d x

y^{^{'}} + 2 y = 7 x y^{2}

\int_{0}^{4} \frac{5 t}{( t - 5 ) ^{2}} d t

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