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

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

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

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

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

ln y^{2} + 1 + C

Solution steps

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

Take the constant out:

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

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

Apply u-substitution

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

= 2 \cdot \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

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

Use the common integral:

\int \frac{1}{u} d u = ln (∣ u ∣)

= 2 \cdot \frac{1}{2} ln ∣ u ∣

Substitute back

u = y^{2} + 1

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

Simplify

2 \cdot \frac{1}{2} ln y^{2} + 1 : ln y^{2} + 1

= ln y^{2} + 1

Add a constant to the solution

= ln y^{2} + 1 + C

x \to 0 - lim (\frac{2}{3} x^{- \frac{1}{3}})

\int_{0}^{\infty} e^{- 14 x} d x

\frac{d ^{3}}{d x ^{3}} (\frac{1 - 5 x}{1 + 5 x})

x \to - 3 lim (14)

f (x) = x^{2} + 8, (2, 12)

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