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

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

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

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

arctan (y) + C

Solution steps

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

Use the common integral:

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

= arctan (y)

Add a constant to the solution

= arctan (y) + C

y = a x 1 - x

\int (5 x - 2) e^{2 x} d x

\int 4 x + 8 d x

\int \frac{200}{t ^{2} + 3 t + 2} d t

\frac{d y}{d x} = x x - 12

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