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

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

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

\int \frac{1}{x ^{2} + π ^{2}} d x

\frac{1}{π} arctan (\frac{x}{π}) + C

Solution steps

\int \frac{1}{x ^{2} + π ^{2}} d x

Apply Integral Substitution

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

Substitute back

u = \frac{x}{π}

= \frac{1}{π} arctan (\frac{x}{π})

Add a constant to the solution

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

\frac{\partial}{\partial x} (3^{s i n (3 x)})

\frac{d}{d x} ((4 x + 3)^{5})

(\frac{2}{s} + \frac{1}{s ^{3}})^{2}

\int_{0}^{9} π (9 - y) d y

f (x) = 49 - x^{2}

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