What is the integral of (x^2)/((1+x^3)^{1.9)} ?

The integral of (x^2)/((1+x^3)^{1.9)} is -(0.37037…)/((1+x^3)^{0.9)}+C

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integral of (x^2)/((1+x^3)^{1.9)}

\int \frac{x ^{2}}{( 1 + x ^{3} ) ^{1.9}} d x

- \frac{0.37037 \dots}{( 1 + x ^{3} ) ^{0.9}} + C

Solution steps

\int \frac{x ^{2}}{( 1 + x ^{3} ) ^{1.9}} d x

Apply u-substitution

: \int \frac{1}{3 u ^{1.9}} d u

= \int \frac{1}{3 u ^{1.9}} 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 ^{1.9}} d u

Apply the Power Rule

: - \frac{1.11111 \dots}{u ^{0.9}}

= \frac{1}{3} (- \frac{1.11111 \dots}{u ^{0.9}})

Substitute back

u = 1 + x^{3}

= \frac{1}{3} (- \frac{1.11111 \dots}{( 1 + x ^{3} ) ^{0.9}})

Simplify

\frac{1}{3} (- \frac{1.11111 \dots}{( 1 + x ^{3} ) ^{0.9}}) : - \frac{0.37037 \dots}{( 1 + x ^{3} ) ^{0.9}}

= - \frac{0.37037 \dots}{( 1 + x ^{3} ) ^{0.9}}

Add a constant to the solution

= - \frac{0.37037 \dots}{( 1 + x ^{3} ) ^{0.9}} + C

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What is the integral of (x^2)/((1+x^3)^{1.9)} ?
The integral of (x^2)/((1+x^3)^{1.9)} is -(0.37037…)/((1+x^3)^{0.9)}+C