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

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

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

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} (cos (\frac{1 - e ^{2 x}}{1 + e ^{2 x}}))

F (r) = \frac{7}{r ^{3}}

\int_{0}^{1} 5 ln (4 x) d x

x + 3, \frac{1}{2} x + 3

\frac{d x}{d t} = x e^{t} - 3 e^{t} + x - 3

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