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

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

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

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

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

Solution steps

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

Apply Integral Substitution

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

Substitute back

u = \frac{x}{b}

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

Add a constant to the solution

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

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