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

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

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

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

arctan (x) + C

Solution steps

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

Simplify

= \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{d}{d x} (\frac{x ^{2} + 100}{x})

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

4 x^{4} - 3 x^{2}, 6 x^{2}

\frac{\partial}{\partial x} (x^{2} y^{3} sin (x z))

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