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

The integral of x/(1+2x^2) is 1/4 ln|1+2x^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+2x^2)

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

\frac{1}{4} ln 1 + 2 x^{2} + C

Solution steps

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

Apply u-substitution

: \int \frac{1}{4 u} d u

= \int \frac{1}{4 u} d u

Take the constant out:

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

= \frac{1}{4} \cdot \int \frac{1}{u} d u

Use the common integral:

\int \frac{1}{u} d u = ln (∣ u ∣)

= \frac{1}{4} ln ∣ u ∣

Substitute back

u = 1 + 2 x^{2}

= \frac{1}{4} ln 1 + 2 x^{2}

Add a constant to the solution

= \frac{1}{4} ln 1 + 2 x^{2} + C

\frac{d}{d x} (x^{5} + 3 x^{2} + 1)

10 x, x^{2} - 11

\frac{d}{d x} (1 - e^{- e^{x}})

y^{^{'}} - 8 y = cos (x)

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

What is the integral of x/(1+2x^2) ?
The integral of x/(1+2x^2) is 1/4 ln|1+2x^2|+C