What is the integral of (x^2)/((1+x^3)^{1.1)} ?

The integral of (x^2)/((1+x^3)^{1.1)} is -(3.33333…)/((1+x^3)^{0.1)}+C

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integral of (x^2)/((1+x^3)^{1.1)}

\int \frac{x ^{2}}{( 1 + x ^{3} ) ^{1.1}} d x

- \frac{3.33333 \dots}{( 1 + x ^{3} ) ^{0.1}} + C

Solution steps

\int \frac{x ^{2}}{( 1 + x ^{3} ) ^{1.1}} d x

Apply u-substitution

: \int \frac{1}{3 u ^{1.1}} d u

= \int \frac{1}{3 u ^{1.1}} 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 ^{1.1}} d u

Apply the Power Rule

: - \frac{10}{u ^{0.1}}

= \frac{1}{3} (- \frac{10}{u ^{0.1}})

Substitute back

u = 1 + x^{3}

= \frac{1}{3} (- \frac{10}{( 1 + x ^{3} ) ^{0.1}})

Simplify

\frac{1}{3} (- \frac{10}{( 1 + x ^{3} ) ^{0.1}}) : - \frac{3.33333 \dots}{( 1 + x ^{3} ) ^{0.1}}

= - \frac{3.33333 \dots}{( 1 + x ^{3} ) ^{0.1}}

Add a constant to the solution

= - \frac{3.33333 \dots}{( 1 + x ^{3} ) ^{0.1}} + C

f (x) = \frac{8 x}{x ^{2} + 1}, at x = 1

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

x \to - 8 lim (\frac{x + 7}{x + 8})

f (x) = ln (2 x + 1), at x = 0

u (x - 5)

What is the integral of (x^2)/((1+x^3)^{1.1)} ?
The integral of (x^2)/((1+x^3)^{1.1)} is -(3.33333…)/((1+x^3)^{0.1)}+C