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

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

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

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

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

ln 1 + u^{2} + C

Solution steps

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

Take the constant out:

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

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

Apply u-substitution

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

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

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}{v} d v

Use the common integral:

\int \frac{1}{v} d v = ln (∣ v ∣)

= 2 \cdot \frac{1}{2} ln ∣ v ∣

Substitute back

v = 1 + u^{2}

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

Simplify

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

= ln 1 + u^{2}

Add a constant to the solution

= ln 1 + u^{2} + C

y^{^{''}} + y^{^{'}} - 2 y = 2 x

\int 2 e^{u} d u

\frac{( 3 y ^{2} - x ^{2} )}{y ^{5}} y^{^{'}} + \frac{x}{2 y ^{4}} = 0

f (x) = - 5.7 + 3.57 ln (x)

\int 1 + cos^{2} (x) d x

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