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

The integral of ((y))/((1+y^{(2)))} is 1/2 ln|1+y^2|+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

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

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

\frac{1}{2} ln 1 + y^{2} + C

Solution steps

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

Apply u-substitution

: \int \frac{1}{2 u} d u

= \int \frac{1}{2 u} d u

Take the constant out:

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

= \frac{1}{2} \cdot \int \frac{1}{u} d u

Use the common integral:

\int \frac{1}{u} d u = ln (∣ u ∣)

= \frac{1}{2} ln ∣ u ∣

Substitute back

u = 1 + y^{2}

= \frac{1}{2} ln 1 + y^{2}

Add a constant to the solution

= \frac{1}{2} ln 1 + y^{2} + C

y^{^{''}} + y^{^{'}} - 6 y = 0, y (0) = 2, y^{^{'}} (0) = - 1

(5 t + 2 y + 6) d t - d y = 0

\frac{d}{d x} (\frac{d y}{d x}) + 1 = 0

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

f (y) = ln (y)

What is the integral of ((y))/((1+y^{(2)))} ?
The integral of ((y))/((1+y^{(2)))} is 1/2 ln|1+y^2|+C