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

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

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

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

\frac{1}{2 5} arctan (\frac{5}{2} x) + C

Solution steps

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

Apply Integral Substitution

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

Take the constant out:

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

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

Use the common integral:

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

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

Substitute back

u = \frac{5}{2} x

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

Add a constant to the solution

= \frac{1}{2 5} arctan (\frac{5}{2} x) + C

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

y^{^{''}} - 4 y^{^{'}} + 4 y = 6 e^{2 t} - 6 t e^{2 t} - 12 t + 12

\int tan^{2} (2 x) sec^{4} (2 x) d x

\frac{d y}{d x} + y = \frac{1}{1 + e ^{x}}

y = (x^{3} - 9 x)^{12}, (3, 0)

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