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

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

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

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

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

Solution steps

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

Apply Integral Substitution

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

Substitute back

u = \frac{x}{t}

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

Add a constant to the solution

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

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f (x) = (x + 9) (x^{2} - 9)

\frac{6}{s ^{4} ( s - 2 ) ^{2}}

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\frac{d}{d x} (2 sec^{2} (5 π x - 1))

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