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

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

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

integral

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

arctan (x) + C

Solution steps

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

Apply exponent rule:

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

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

Use the common integral:

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

= arctan (x)

Add a constant to the solution

= arctan (x) + C

\frac{e ^{x}}{( x + 2 )}

e^{x} cos^{2} (x)

y = cos (x^{2}) (0.1)

t^{2} (e^{t})^{2}

4 x^{10}

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