What is the integral of 1/(x^2+6) ?

The integral of 1/(x^2+6) is 1/(sqrt(6))arctan(x/(sqrt(6)))+C

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integral of 1/(x^2+6)

\int \frac{1}{x ^{2} + 6} d x

\frac{1}{6} arctan (\frac{x}{6}) + C

Solution steps

\int \frac{1}{x ^{2} + 6} d x

Apply Integral Substitution

= \int \frac{1}{6 ( 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}{6} \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}{6} arctan (u)

Substitute back

u = \frac{x}{6}

= \frac{1}{6} arctan (\frac{x}{6})

Add a constant to the solution

= \frac{1}{6} arctan (\frac{x}{6}) + C

\int t (t^{2} - 1) d t

\frac{d}{d x} (3 e^{(x^{2})})

\int cos (x^{3}) d x

y = 7 x + 11, (2, 5)

e^{3 a - b t}

What is the integral of 1/(x^2+6) ?
The integral of 1/(x^2+6) is 1/(sqrt(6))arctan(x/(sqrt(6)))+C