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

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

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integral of 1/(3+x^2)

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

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

Solution steps

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

Apply Integral Substitution

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

Substitute back

u = \frac{x}{3}

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

Add a constant to the solution

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

x \to 3 lim (\frac{x + 1}{x - 2})

g (x) = 15 x

\frac{d}{d x} (14)

\int_{1}^{3} x d x

\frac{\partial}{\partial x} (e^{- x} sec (y))

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