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

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

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

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

- ln 1 - x^{4} + C

Solution steps

\int \frac{4 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

= 4 \cdot \int \frac{x ^{3}}{1 - x ^{4}} 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 (- \frac{1}{4} \cdot \int \frac{1}{u} d u)

Use the common integral:

\int \frac{1}{u} d u = ln (∣ u ∣)

= 4 (- \frac{1}{4} ln ∣ u ∣)

Substitute back

u = 1 - x^{4}

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

Simplify

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

= - ln 1 - x^{4}

Add a constant to the solution

= - ln 1 - x^{4} + C

\int x^{5} 6 x^{6} + 2 d x

\int (4 x^{2} + 3)^{2} d x

x^{2} - x, 2 x - 2 x^{2}

x \to 4 lim (\frac{1}{x + 2})

e^{x} - 6 x^{5}

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