What is the integral of 4/(1+y^2) ?

The integral of 4/(1+y^2) is 4arctan(y)+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral of 4/(1+y^2)

\int \frac{4}{1 + y ^{2}} d y

4 arctan (y) + C

Solution steps

\int \frac{4}{1 + y ^{2}} d y

Take the constant out:

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

= 4 \cdot \int \frac{1}{1 + y ^{2}} d y

Use the common integral:

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

= 4 arctan (y)

Add a constant to the solution

= 4 arctan (y) + C

\int 1 + \frac{4}{9} x^{- \frac{2}{3}} d x

\int x (1 + x)^{3} d x

\int t^{5} sin (t^{6}) d t

\int \frac{x ^{7} + x ^{2} + 1}{x ^{2}} d x

\int \frac{8}{x ^{5}} d x