What is the integral of (x^2)/((6-x^3)^5) ?

The integral of (x^2)/((6-x^3)^5) is 1/(12(6-x^3)^4)+C

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integral of (x^2)/((6-x^3)^5)

\int \frac{x ^{2}}{( 6 - x ^{3} ) ^{5}} d x

\frac{1}{12 ( 6 - x ^{3} ) ^{4}} + C

Solution steps

\int \frac{x ^{2}}{( 6 - x ^{3} ) ^{5}} d x

Apply u-substitution

: \int \frac{u}{2 x ( 6 - u ^{\frac{3}{2}} ) ^{5}} d u

= \int \frac{1}{3 ( 6 - u ) ^{5}} d u

Take the constant out:

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

= \frac{1}{3} \cdot \int \frac{1}{( 6 - u ) ^{5}} d u

Apply u-substitution

: \int - \frac{1}{v ^{5}} d v

= \frac{1}{3} \cdot \int - \frac{1}{v ^{5}} d v

Take the constant out:

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

= \frac{1}{3} (- \int \frac{1}{v ^{5}} d v)

Apply the Power Rule

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

= \frac{1}{3} (- (- \frac{1}{4 v ^{4}}))

Substitute back

= \frac{1}{3} (- (- \frac{1}{4 ( 6 - x ^{3} ) ^{4}}))

Simplify

\frac{1}{3} (- (- \frac{1}{4 ( 6 - x ^{3} ) ^{4}})) : \frac{1}{12 ( 6 - x ^{3} ) ^{4}}

= \frac{1}{12 ( 6 - x ^{3} ) ^{4}}

Add a constant to the solution

= \frac{1}{12 ( 6 - x ^{3} ) ^{4}} + C

y = ln (\frac{x}{x ^{2} + 1})

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

\int \frac{1}{( x ^{2} - 25 ) ^{2}} d x

f (x) = x^{2} + 5, at x = 5

x \to 3 lim (2)

What is the integral of (x^2)/((6-x^3)^5) ?
The integral of (x^2)/((6-x^3)^5) is 1/(12(6-x^3)^4)+C