What is the integral of (48x)/(4x^2+e) ?

The integral of (48x)/(4x^2+e) is 6ln|4x^2+e|+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral of (48x)/(4x^2+e)

\int \frac{48 x}{4 x ^{2} + e} d x

6 ln 4 x^{2} + e + C

Solution steps

\int \frac{48 x}{4 x ^{2} + e} d x

Take the constant out:

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

= 48 \cdot \int \frac{x}{4 x ^{2} + e} d x

Apply u-substitution

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

= 48 \cdot \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

= 48 \cdot \frac{1}{8} \cdot \int \frac{1}{u} d u

Use the common integral:

\int \frac{1}{u} d u = ln (∣ u ∣)

= 48 \cdot \frac{1}{8} ln ∣ u ∣

Substitute back

u = 4 x^{2} + e

= 48 \cdot \frac{1}{8} ln 4 x^{2} + e

Simplify

48 \cdot \frac{1}{8} ln 4 x^{2} + e : 6 ln 4 x^{2} + e

= 6 ln 4 x^{2} + e

Add a constant to the solution

= 6 ln 4 x^{2} + e + C

\int_{- 1}^{1} \frac{x}{x ^{2} + 1} d x

\int 5 x^{\frac{2}{3}} d x

y = 6 x^{2} - x^{3}, (1, 5)

n = 0 \sum \infty \frac{1}{1 !}

\int 5^{x + 2} d x

What is the integral of (48x)/(4x^2+e) ?
The integral of (48x)/(4x^2+e) is 6ln|4x^2+e|+C