What is the integral of (\sqrt[3]{x^2}-1/(x^2)) ?

The integral of (\sqrt[3]{x^2}-1/(x^2)) is 3/5 x^{5/3}+1/x+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral of (\sqrt[3]{x^2}-1/(x^2))

\frac{3}{5} x^{\frac{5}{3}} + \frac{1}{x} + C

Solution steps

Apply the Sum Rule:

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

\int \frac{1}{x ^{2}} d x = - \frac{1}{x}

= \frac{3}{5} x^{\frac{5}{3}} - (- \frac{1}{x})

Simplify

= \frac{3}{5} x^{\frac{5}{3}} + \frac{1}{x}

Add a constant to the solution

= \frac{3}{5} x^{\frac{5}{3}} + \frac{1}{x} + C

\frac{d}{d x} (6 x^{4} - 7 x^{3} + 2 x + 2)

\int \frac{z}{x ^{2} + z ^{2}} d x

\int (sin (x))^{2} (cos (x)) d x

\frac{d}{d x} (6 x^{\frac{7}{2}} + 4 x^{\frac{5}{2}} + 2 x)

What is the integral of (\sqrt[3]{x^2}-1/(x^2)) ?
The integral of (\sqrt[3]{x^2}-1/(x^2)) is 3/5 x^{5/3}+1/x+C