What is the integral of (x^3)/((6-x^4)^2) ?

The integral of (x^3)/((6-x^4)^2) is 1/(4(6-x^4))+C

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integral of (x^3)/((6-x^4)^2)

\int \frac{x ^{3}}{( 6 - x ^{4} ) ^{2}} d x

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

Solution steps

\int \frac{x ^{3}}{( 6 - x ^{4} ) ^{2}} d x

Apply u-substitution

: \int - \frac{1}{4 u ^{2}} d u

= \int - \frac{1}{4 u ^{2}} 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 ^{2}} d u

Apply the Power Rule

: - \frac{1}{u}

= - \frac{1}{4} (- \frac{1}{u})

Substitute back

u = 6 - x^{4}

= - \frac{1}{4} (- \frac{1}{6 - x ^{4}})

Simplify

- \frac{1}{4} (- \frac{1}{6 - x ^{4}}) : \frac{1}{4 ( 6 - x ^{4} )}

= \frac{1}{4 ( 6 - x ^{4} )}

Add a constant to the solution

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

\int \frac{7 x ^{2} + 16 x - 19}{x ^{2} + 2 x - 3} d x

f (x) = \frac{1}{3 ( x ) ^{2} + 2 ( x )}

f (x) = 2 x^{3} + 42 x^{2} + 288 x + 14

y = 2 - 6 x

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

What is the integral of (x^3)/((6-x^4)^2) ?
The integral of (x^3)/((6-x^4)^2) is 1/(4(6-x^4))+C