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

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

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

integral

\frac{1}{( 1 + a ^{2} t ^{2} )}

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

Solution steps

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

Apply Integral Substitution

= \int \frac{1}{t ( 1 + u ^{2} )} 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}{1 + u ^{2}} d u

Use the common integral:

\int \frac{1}{1 + u ^{2}} d u = arctan (u)

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

Substitute back

u = t a

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

Add a constant to the solution

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

(3^{x} + 3^{- x})^{2}

(\frac{1 - 1}{n})^{n - 1}

x^{2} ln (x + 1)

x (1 + x^{2})^{15}

\frac{x}{x ^{2} + y ^{2}}

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