What is the integral of (16x^3-4x)/(2x^4-x^2+3) ?

The integral of (16x^3-4x)/(2x^4-x^2+3) is 2ln|2x^4-x^2+3|+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral of (16x^3-4x)/(2x^4-x^2+3)

\int \frac{16 x ^{3} - 4 x}{2 x ^{4} - x ^{2} + 3} d x

2 ln 2 x^{4} - x^{2} + 3 + C

Solution steps

\int \frac{16 x ^{3} - 4 x}{2 x ^{4} - x ^{2} + 3} d x

Apply u-substitution

: \int \frac{2}{u} d u

= \int \frac{2}{u} d u

Take the constant out:

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

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

Use the common integral:

\int \frac{1}{u} d u = ln (∣ u ∣)

= 2 ln ∣ u ∣

Substitute back

u = 2 x^{4} - x^{2} + 3

= 2 ln 2 x^{4} - x^{2} + 3

Add a constant to the solution

= 2 ln 2 x^{4} - x^{2} + 3 + C

x \to 0 lim (sin (\frac{3}{x}))

\int 5 e^{6 x} d x

\int (10 - 5 x) d x

x^{2} y^{^{''}} + 3 x y^{^{'}} + 5 y = 0

\int sin^{5} (4 x) cos (4 x) d x

What is the integral of (16x^3-4x)/(2x^4-x^2+3) ?
The integral of (16x^3-4x)/(2x^4-x^2+3) is 2ln|2x^4-x^2+3|+C