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

The integral of ((x^2+1)(x^2-2))/(x^{2/3)} is 3/13 x^{13/3}-3/7 x^{7/3}-6x^{1/3}+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

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

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

\frac{3}{13} x^{\frac{13}{3}} - \frac{3}{7} x^{\frac{7}{3}} - 6 x^{\frac{1}{3}} + C

Solution steps

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

Expand

\frac{( x ^{2} + 1 ) ( x ^{2} - 2 )}{x ^{\frac{2}{3}}} : x^{\frac{10}{3}} - x^{\frac{4}{3}} - \frac{2}{x ^{\frac{2}{3}}}

= \int x^{\frac{10}{3}} - x^{\frac{4}{3}} - \frac{2}{x ^{\frac{2}{3}}} d x

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= \int x^{\frac{10}{3}} d x - \int x^{\frac{4}{3}} d x - \int \frac{2}{x ^{\frac{2}{3}}} d x

\int x^{\frac{10}{3}} d x = \frac{3}{13} x^{\frac{13}{3}}

\int x^{\frac{4}{3}} d x = \frac{3}{7} x^{\frac{7}{3}}

\int \frac{2}{x ^{\frac{2}{3}}} d x = 6 x^{\frac{1}{3}}

= \frac{3}{13} x^{\frac{13}{3}} - \frac{3}{7} x^{\frac{7}{3}} - 6 x^{\frac{1}{3}}

Add a constant to the solution

= \frac{3}{13} x^{\frac{13}{3}} - \frac{3}{7} x^{\frac{7}{3}} - 6 x^{\frac{1}{3}} + C

a re a y = x + 2, x = - 2, y = 2

t an g e n t f (x) = 11 x^{2} + 6 x, at x = 2

x \to - 5 - lim (4 x + 25)

\int \frac{r}{1 + r ^{2}} d r

\int \frac{1}{y ^{2} + 2} d y

What is the integral of ((x^2+1)(x^2-2))/(x^{2/3)} ?
The integral of ((x^2+1)(x^2-2))/(x^{2/3)} is 3/13 x^{13/3}-3/7 x^{7/3}-6x^{1/3}+C