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

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

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integral of 1/((100+x^2))

integral

\frac{1}{( 100 + x ^{2} )}

\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

\frac{( r ^{2} )}{( 1 + r ^{2} ) d}

sin^{3} (2 x) cos (2 x)

(1 + 8 x)^{\frac{3}{2}}

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

\frac{( e ^{x} )}{x ^{- 2}}

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