What is the integral of (x^3)/(6-x^4) ?

The integral of (x^3)/(6-x^4) is -1/4 ln|6-x^4|+C

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integral of (x^3)/(6-x^4)

\int \frac{x ^{3}}{6 - x ^{4}} d x

- \frac{1}{4} ln 6 - x^{4} + C

Solution steps

\int \frac{x ^{3}}{6 - x ^{4}} d x

Apply u-substitution

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

= \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

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

Use the common integral:

\int \frac{1}{u} d u = ln (∣ u ∣)

= - \frac{1}{4} ln ∣ u ∣

Substitute back

u = 6 - x^{4}

= - \frac{1}{4} ln 6 - x^{4}

Add a constant to the solution

= - \frac{1}{4} ln 6 - x^{4} + C

\frac{6 u ^{2}}{( u ^{2} + u ) ^{3}}

[δ (t + 1)]

\int \frac{- x + 2}{x ^{2} - x + 1} d x

\int sech (\frac{1}{4}) x tanh (\frac{1}{4}) x d x

\frac{3 s ^{2}}{( s ^{2} + 4 ) ^{2}}

What is the integral of (x^3)/(6-x^4) ?
The integral of (x^3)/(6-x^4) is -1/4 ln|6-x^4|+C