What is the integral of (x^2)/((16-x^3)^2) ?

The integral of (x^2)/((16-x^3)^2) is 1/(3(16-x^3))+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)/((16-x^3)^2)

\int \frac{x ^{2}}{( 16 - x ^{3} ) ^{2}} d x

\frac{1}{3 ( 16 - x ^{3} )} + C

Solution steps

\int \frac{x ^{2}}{( 16 - x ^{3} ) ^{2}} d x

Apply u-substitution

: \int - \frac{1}{3 u ^{2}} d u

= \int - \frac{1}{3 u ^{2}} 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 ^{2}} d u

Apply the Power Rule

: - \frac{1}{u}

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

Substitute back

u = 16 - x^{3}

= - \frac{1}{3} (- \frac{1}{16 - x ^{3}})

Simplify

- \frac{1}{3} (- \frac{1}{16 - x ^{3}}) : \frac{1}{3 ( 16 - x ^{3} )}

= \frac{1}{3 ( 16 - x ^{3} )}

Add a constant to the solution

= \frac{1}{3 ( 16 - x ^{3} )} + C

x \to - 4 + lim (\frac{x + 3}{x + 4})

\int x^{2} sec^{2} (x) tan (x) d x

y = x^{2} - 4, y = 2 - 5 x^{2}

x \to 0 lim (\frac{6 sin ( 2 x )}{x})

(2, - 14), (1, - 11)

What is the integral of (x^2)/((16-x^3)^2) ?
The integral of (x^2)/((16-x^3)^2) is 1/(3(16-x^3))+C