What is the integral of (x^3)/((7-x^4)^5) ?

The integral of (x^3)/((7-x^4)^5) is 1/(16(7-x^4)^4)+C

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integral of (x^3)/((7-x^4)^5)

\int \frac{x ^{3}}{( 7 - x ^{4} ) ^{5}} d x

\frac{1}{16 ( 7 - x ^{4} ) ^{4}} + C

Solution steps

\int \frac{x ^{3}}{( 7 - x ^{4} ) ^{5}} d x

Apply u-substitution

: \int - \frac{1}{4 u ^{5}} d u

= \int - \frac{1}{4 u ^{5}} 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 ^{5}} d u

Apply the Power Rule

: - \frac{1}{4 u ^{4}}

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

Substitute back

u = 7 - x^{4}

= - \frac{1}{4} (- \frac{1}{4 ( 7 - x ^{4} ) ^{4}})

Simplify

- \frac{1}{4} (- \frac{1}{4 ( 7 - x ^{4} ) ^{4}}) : \frac{1}{16 ( 7 - x ^{4} ) ^{4}}

= \frac{1}{16 ( 7 - x ^{4} ) ^{4}}

Add a constant to the solution

= \frac{1}{16 ( 7 - x ^{4} ) ^{4}} + C

\frac{\partial}{\partial x} (e^{x (e^{y})})

2 (x - 2) - cos (3 x) \cdot 3

\int - e^{- t} d t

\frac{d}{d x} (arcsec (2 x))

x \to 2 lim (x^{2} + 3)

What is the integral of (x^3)/((7-x^4)^5) ?
The integral of (x^3)/((7-x^4)^5) is 1/(16(7-x^4)^4)+C