What is the integral of 1/((4+9x^2) 5/2) ?

The integral of 1/((4+9x^2) 5/2) is 1/15 arctan(3/2 x)+C

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integral of 1/((4+9x^2) 5/2)

\int \frac{1}{( 4 + 9 x ^{2} ) \frac{5}{2}} d x

\frac{1}{15} arctan (\frac{3}{2} x) + C

Solution steps

\int \frac{1}{( 4 + 9 x ^{2} ) \frac{5}{2}} d x

Take the constant out:

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

= \frac{1}{\frac{5}{2}} \cdot \int \frac{1}{4 + 9 x ^{2}} d x

Simplify

= \frac{2}{5} \cdot \int \frac{1}{4 + 9 x ^{2}} d x

Apply Integral Substitution

= \frac{2}{5} \cdot \int \frac{1}{6 ( u ^{2} + 1 )} d u

Take the constant out:

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

= \frac{2}{5} \cdot \frac{1}{6} \cdot \int \frac{1}{u ^{2} + 1} d u

Use the common integral:

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

= \frac{2}{5} \cdot \frac{1}{6} arctan (u)

Substitute back

u = \frac{3}{2} x

= \frac{2}{5} \cdot \frac{1}{6} arctan (\frac{3}{2} x)

Simplify

\frac{2}{5} \cdot \frac{1}{6} arctan (\frac{3}{2} x) : \frac{1}{15} arctan (\frac{3}{2} x)

= \frac{1}{15} arctan (\frac{3}{2} x)

Add a constant to the solution

= \frac{1}{15} arctan (\frac{3}{2} x) + C

\int \frac{x}{3} d x

\int [(4 x - 9)^{6} - 2 x] d x

\int \frac{3 x}{1 + x ^{4}} d x

y^{^{'}} + y^{3} = 0, y (0) = 3

\frac{\partial}{\partial x} (x^{3} y^{4} z^{2})

What is the integral of 1/((4+9x^2) 5/2) ?
The integral of 1/((4+9x^2) 5/2) is 1/15 arctan(3/2 x)+C