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

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

\frac{d}{d x} (e^{y} + \frac{y}{x})

x \to - 2 lim (\frac{2 x ^{2} + 3 x}{x - 3})

\frac{d}{d x} (arcsin (8 x) + arccos (8 x))

y^{^{''}} - 6 y^{^{'}} - 16 y = 0

\frac{d}{d x} (4 x^{0.75} \cdot (240 - 2 x)^{0.5})