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

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

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

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

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

Solution steps

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

Apply Integral Substitution

= \int \frac{1}{3 ( 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}{3} \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}{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

\frac{d}{d x} (\frac{3 x}{2 sin ( x ) + cos ( x )})

y^{^{''}} + y^{^{'}} - 6 y = 0

y = x^{3} - 9 x^{2} + 14 x, y = - x^{3} + 9 x^{2} - 14 x

n = 1 \sum \infty \frac{5 ^{n}}{6 ^{n}}

\int_{0}^{0.6} 2000 x d x

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