What is the integral of x/(2+4x^2) ?

The integral of x/(2+4x^2) is 1/8 ln|2+4x^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/(2+4x^2)

\int \frac{x}{2 + 4 x ^{2}} d x

\frac{1}{8} ln 2 + 4 x^{2} + C

Solution steps

\int \frac{x}{2 + 4 x ^{2}} d x

Apply u-substitution

: \int \frac{1}{8 u} d u

= \int \frac{1}{8 u} d u

Take the constant out:

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

= \frac{1}{8} \cdot \int \frac{1}{u} d u

Use the common integral:

\int \frac{1}{u} d u = ln (∣ u ∣)

= \frac{1}{8} ln ∣ u ∣

Substitute back

u = 2 + 4 x^{2}

= \frac{1}{8} ln 2 + 4 x^{2}

Add a constant to the solution

= \frac{1}{8} ln 2 + 4 x^{2} + C

\int - 15 cos^{3} (t) sin (t) d t

\int arccos (a) x d x

\int \frac{r}{r ^{2} - x ^{2}} d x

\int 3 cos^{4} (x) sin (x) d x

\int \frac{2 x ^{3} + 10 x}{x ^{4} + 2 x ^{2} + 1} d x

What is the integral of x/(2+4x^2) ?
The integral of x/(2+4x^2) is 1/8 ln|2+4x^2|+C