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

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

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

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

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

Solution steps

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

Apply Integral Substitution

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

Substitute back

u = 2 x

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

Add a constant to the solution

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

(x + y + 4) d x + (2 x + 2 y + 1) d y = 0

\frac{d y}{d x} (25 - x^{2}) + 5 x y = 0

\frac{d}{d x} (\frac{( 1 - x ) ^{2}}{x ^{2}})

y^{^{'}} = y x e^{x^{2}} - y

x \to 0 - lim (x^{2} ln (x))

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