What is the integral of (x^2)/((1+x^3)^{1.7)} ?

The integral of (x^2)/((1+x^3)^{1.7)} is -(0.47619…)/((1+x^3)^{0.7)}+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral of (x^2)/((1+x^3)^{1.7)}

\int \frac{x ^{2}}{( 1 + x ^{3} ) ^{1.7}} d x

- \frac{0.47619 \dots}{( 1 + x ^{3} ) ^{0.7}} + C

Solution steps

\int \frac{x ^{2}}{( 1 + x ^{3} ) ^{1.7}} d x

Apply u-substitution

: \int \frac{1}{3 u ^{1.7}} d u

= \int \frac{1}{3 u ^{1.7}} 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 ^{1.7}} d u

Apply the Power Rule

: - \frac{1.42857 \dots}{u ^{0.7}}

= \frac{1}{3} (- \frac{1.42857 \dots}{u ^{0.7}})

Substitute back

u = 1 + x^{3}

= \frac{1}{3} (- \frac{1.42857 \dots}{( 1 + x ^{3} ) ^{0.7}})

Simplify

\frac{1}{3} (- \frac{1.42857 \dots}{( 1 + x ^{3} ) ^{0.7}}) : - \frac{0.47619 \dots}{( 1 + x ^{3} ) ^{0.7}}

= - \frac{0.47619 \dots}{( 1 + x ^{3} ) ^{0.7}}

Add a constant to the solution

= - \frac{0.47619 \dots}{( 1 + x ^{3} ) ^{0.7}} + C

(38) (3 - 4)

x \to \infty lim (e^{- 3 x} sin (x))

\int_{- 1}^{1} (x^{2} - x^{4}) d x

- 4 sin (x) - 9 ln (x) + 11

What is the integral of (x^2)/((1+x^3)^{1.7)} ?
The integral of (x^2)/((1+x^3)^{1.7)} is -(0.47619…)/((1+x^3)^{0.7)}+C