What is the integral of sqrt(x)-1/(x^2) ?

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

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

\frac{2}{3} x^{\frac{3}{2}} + \frac{1}{x} + C

Solution steps

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

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

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

= \frac{2}{3} x^{\frac{3}{2}} - (- \frac{1}{x})

Simplify

= \frac{2}{3} x^{\frac{3}{2}} + \frac{1}{x}

Add a constant to the solution

= \frac{2}{3} x^{\frac{3}{2}} + \frac{1}{x} + C

\int e^{8 x} cos (4 x) d x

\int \frac{1}{1 + 2 t} d t

\int e^{3 l n (x)} + e^{3 x} d x

\int (3 x - 1)^{6} d x

\int \frac{x}{a ^{2} + b ^{2} x ^{2}} d x

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