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

The integral of ((x^2+x))/((4-3x^2-2x^3)^4) is 1/(18(4-3x^2-2x^3)^3)+C

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

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

\frac{1}{18 ( 4 - 3 x ^{2} - 2 x ^{3} ) ^{3}} + C

Solution steps

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

Apply u-substitution

: \int - \frac{1}{6 u ^{4}} d u

= \int - \frac{1}{6 u ^{4}} d u

Take the constant out:

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

= - \frac{1}{6} \cdot \int \frac{1}{u ^{4}} d u

Apply the Power Rule

: - \frac{1}{3 u ^{3}}

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

Substitute back

u = 4 - 3 x^{2} - 2 x^{3}

= - \frac{1}{6} (- \frac{1}{3 ( 4 - 3 x ^{2} - 2 x ^{3} ) ^{3}})

Simplify

- \frac{1}{6} (- \frac{1}{3 ( 4 - 3 x ^{2} - 2 x ^{3} ) ^{3}}) : \frac{1}{18 ( 4 - 3 x ^{2} - 2 x ^{3} ) ^{3}}

= \frac{1}{18 ( 4 - 3 x ^{2} - 2 x ^{3} ) ^{3}}

Add a constant to the solution

= \frac{1}{18 ( 4 - 3 x ^{2} - 2 x ^{3} ) ^{3}} + C

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What is the integral of ((x^2+x))/((4-3x^2-2x^3)^4) ?
The integral of ((x^2+x))/((4-3x^2-2x^3)^4) is 1/(18(4-3x^2-2x^3)^3)+C