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

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

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integral of (-1)/(y^2+1)

\int \frac{- 1}{y ^{2} + 1} d y

- arctan (y) + C

Solution steps

\int \frac{- 1}{y ^{2} + 1} d y

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= - \int \frac{1}{y ^{2} + 1} d y

Use the common integral:

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

= - arctan (y)

Add a constant to the solution

= - arctan (y) + C

y^{^{''}} + 2 y = sin (2 t), y (0) = 0, y^{^{'}} (0) = 0

\int \frac{1}{sec ( x ) + tan ( x )} d x

\frac{1}{s ^{3} + 8 s ^{2} + 12 s}

\frac{d}{d x} (2 x^{- 3 x})

f (x) = 4 x^{\frac{4}{3}} - \frac{2}{3} x^{\frac{3}{2}} + x^{2} - 4 x + 1

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