What is the integral of 1/(81+x^2) ?

The integral of 1/(81+x^2) is 1/9 arctan(x/9)+C

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integral of 1/(81+x^2)

\int \frac{1}{81 + x ^{2}} d x

\frac{1}{9} arctan (\frac{x}{9}) + C

Solution steps

\int \frac{1}{81 + x ^{2}} d x

Apply Integral Substitution

= \int \frac{1}{9 ( 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}{9} \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}{9} arctan (u)

Substitute back

u = \frac{x}{9}

= \frac{1}{9} arctan (\frac{x}{9})

Add a constant to the solution

= \frac{1}{9} arctan (\frac{x}{9}) + C

x \to 0 lim (0.1 ln (0.1))

\int sin^{2} (x) (cos (x)) d x

\frac{7 x - 2}{9 x + 1}

\frac{\partial}{\partial x} (x \cdot ln (t))

\frac{d v}{d t} = 9.81 - \frac{0.5}{v ^{- 1}}

What is the integral of 1/(81+x^2) ?
The integral of 1/(81+x^2) is 1/9 arctan(x/9)+C