What is the integral of-(x^2-2x^5)(x^3-x^6)^{-10} ?

The integral of-(x^2-2x^5)(x^3-x^6)^{-10} is 1/(27(x^3-x^6)^9)+C

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integral of-(x^2-2x^5)(x^3-x^6)^{-10}

\int - (x^{2} - 2 x^{5}) (x^{3} - x^{6})^{- 10} d x

\frac{1}{27 ( x ^{3} - x ^{6} ) ^{9}} + C

Solution steps

\int - (x^{2} - 2 x^{5}) (x^{3} - x^{6})^{- 10} d x

Take the constant out:

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

= - \int (x^{2} - 2 x^{5}) (x^{3} - x^{6})^{- 10} d x

Apply u-substitution

= - \int \frac{1}{3 u ^{10}} 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}{u ^{10}} d u

Apply the Power Rule

= - \frac{1}{3} (- \frac{1}{9 u ^{9}})

Substitute back

u = x^{3} - x^{6}

= - \frac{1}{3} (- \frac{1}{9 ( x ^{3} - x ^{6} ) ^{9}})

Simplify

- \frac{1}{3} (- \frac{1}{9 ( x ^{3} - x ^{6} ) ^{9}}) : \frac{1}{27 ( x ^{3} - x ^{6} ) ^{9}}

= \frac{1}{27 ( x ^{3} - x ^{6} ) ^{9}}

Add a constant to the solution

= \frac{1}{27 ( x ^{3} - x ^{6} ) ^{9}} + C

What is the integral of-(x^2-2x^5)(x^3-x^6)^{-10} ?
The integral of-(x^2-2x^5)(x^3-x^6)^{-10} is 1/(27(x^3-x^6)^9)+C