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)

integral

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

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

\frac{( x )}{( 1 + x ^{\frac{1}{4}} )}

\frac{x ^{2}}{9}

\frac{e ^{- x}}{2}

\frac{6}{( 1 + t ^{2} )}

a^{2} + x^{2}