What is the integral of 1/(x^2+8) ?

The integral of 1/(x^2+8) is 1/(2sqrt(2))arctan(x/(2sqrt(2)))+C

integral of 1/(x^2+8)

\int \frac{1}{x ^{2} + 8} d x

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

Solution steps

\int \frac{1}{x ^{2} + 8} d x

Apply Integral Substitution

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

Substitute back

u = \frac{x}{2 2}

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

Add a constant to the solution

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

\int \frac{1}{x 9 x ^{2} + 1} d x

\int (x^{8}) d x

t an g e n t f (x) = x^{3} - 9 x, (3, 0)

t an g e n t f (x) = \frac{4 x}{x + 2}, (6, 3)

d er i v a t i v e f (x) = x^{2} + 5 x

What is the integral of 1/(x^2+8) ?
The integral of 1/(x^2+8) is 1/(2sqrt(2))arctan(x/(2sqrt(2)))+C