What is the integral of (1+t^2)^{-1} ?

The integral of (1+t^2)^{-1} is arctan(t)+C

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integral of (1+t^2)^{-1}

integral

(1 + t^{2})^{- 1}

arctan (t) + C

Solution steps

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

Apply exponent rule:

a^{- 1} = \frac{1}{a}

= \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{( 7 ^{x} - 1 )}{x}

\frac{x}{4 + x}

\frac{1}{( x ^{2} + 2 x + 4 ) ^{\frac{5}{2}}}

\frac{s}{( s ^{2} + a ^{2} )}

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

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