What is the integral of 1/(100+x^2) ?

The integral of 1/(100+x^2) is 1/10 arctan(x/(10))+C

integral of 1/(100+x^2)

\int \frac{1}{100 + x ^{2}} d x

\frac{1}{10} arctan (\frac{x}{10}) + C

Solution steps

\int \frac{1}{100 + x ^{2}} d x

Apply Integral Substitution

= \int \frac{1}{10 ( 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}{10} \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}{10} arctan (u)

Substitute back

u = \frac{x}{10}

= \frac{1}{10} arctan (\frac{x}{10})

Add a constant to the solution

= \frac{1}{10} arctan (\frac{x}{10}) + C

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What is the integral of 1/(100+x^2) ?
The integral of 1/(100+x^2) is 1/10 arctan(x/(10))+C