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

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

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

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

ln x^{4} - 2 + C

Solution steps

\int \frac{4 x ^{3}}{x ^{4} - 2} 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} - 2} 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} - 2

= 4 \cdot \frac{1}{4} ln x^{4} - 2

Simplify

4 \cdot \frac{1}{4} ln x^{4} - 2 : ln x^{4} - 2

= ln x^{4} - 2

Add a constant to the solution

= ln x^{4} - 2 + C

\frac{\partial}{\partial x} (cos (x^{2} + y^{2}) \cdot 2 x)

\int_{0}^{π} cos^{4} (2 t) d t

\frac{d}{d x} ((x^{3} - x)^{3})

x^{2} y^{^{'}} + 2 x y = sin (3 x)

x \to 0 - lim (\frac{x + 94}{x})

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