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

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

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral of (4x^3)/(x^4-3)

\int \frac{4 x ^{3}}{x ^{4} - 3} d x

ln x^{4} - 3 + C

Solution steps

\int \frac{4 x ^{3}}{x ^{4} - 3} d x

Take the constant out:

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

= 4 \cdot \int \frac{x ^{3}}{x ^{4} - 3} d x

Apply u-substitution

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

= 4 \cdot \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

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

Use the common integral:

\int \frac{1}{u} d u = ln (∣ u ∣)

= 4 \cdot \frac{1}{4} ln ∣ u ∣

Substitute back

u = x^{4} - 3

= 4 \cdot \frac{1}{4} ln x^{4} - 3

Simplify

4 \cdot \frac{1}{4} ln x^{4} - 3 : ln x^{4} - 3

= ln x^{4} - 3

Add a constant to the solution

= ln x^{4} - 3 + C

\int sin^{2} (3 u) d u

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

\int \frac{1}{- e ^{x}} d x

\int \frac{1 + w}{3 - w ^{2}} d w

\int \frac{3}{1 - cos ( x )} d x

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