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

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

\int \frac{( x ^{2} - 2 ) ( x ^{2} + 2 )}{4 x ^{3}} d x

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

\int x + x^{- 1} d x

\int \frac{3 x}{1 + x ^{2}} d x

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