What is the integral of 1/(3+4x^2) ?

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

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integral of 1/(3+4x^2)

\int \frac{1}{3 + 4 x ^{2}} d x

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

Solution steps

\int \frac{1}{3 + 4 x ^{2}} d x

Apply Integral Substitution

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

Substitute back

u = \frac{2}{3} x

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

Add a constant to the solution

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

y = e^{x + 2}

n = 4 \sum \infty (\frac{3}{4})^{- n}

\frac{d y}{d x} + y \cdot (x^{2} - 1) + x \cdot y^{6} = 0

\frac{d}{d x} (\frac{15 x ^{7} + 27 x ^{5}}{3 x})

y = x^{3 x}

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