What is the integral of (2x^3)/(1-x^4) ?

The integral of (2x^3)/(1-x^4) is -1/2 ln|1-x^4|+C

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integral of (2x^3)/(1-x^4)

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

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

Solution steps

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

Take the constant out:

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

= 2 \cdot \int \frac{x ^{3}}{1 - x ^{4}} d x

Apply u-substitution

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

= 2 \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

= 2 (- \frac{1}{4} \cdot \int \frac{1}{u} d u)

Use the common integral:

\int \frac{1}{u} d u = ln (∣ u ∣)

= 2 (- \frac{1}{4} ln ∣ u ∣)

Substitute back

u = 1 - x^{4}

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

Simplify

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

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

Add a constant to the solution

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

\int \frac{1}{( 1 - 4 x ) ^{\frac{3}{2}}} d x

\int \frac{( x ^{4} )}{4} d x

\int \frac{1}{x ( 2 + 3 ln ( x ) )} d x

\int \frac{1}{16 - 9 x ^{2}} d x

\int t^{4} + t^{2} d t

What is the integral of (2x^3)/(1-x^4) ?
The integral of (2x^3)/(1-x^4) is -1/2 ln|1-x^4|+C