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

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

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

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

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

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

Solution steps

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

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 + x^{2}

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

Add a constant to the solution

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

\int e^{6 x} cos (e^{3 x}) d x

\int \frac{x + 3}{x ^{2} + 3 x + 2} d x

\int x e^{- 14 x} d x

\int sin (15 x) sin (6 x) d x

\int cos (2 x) cot (2 x) d x

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