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

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

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

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

arctan (t) + C

Solution steps

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

Use the common integral:

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

= arctan (t)

Add a constant to the solution

= arctan (t) + C

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