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

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

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

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

Solution steps

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

Expand

\frac{2 x ^{2} + x ^{2} x - 1}{x ^{2}} : 2 + x - \frac{1}{x ^{2}}

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

\int 2 d x = 2 x

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

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

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

Simplify

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

Add a constant to the solution

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

\int \frac{x + 6}{2 x ^{2} + x - 1} d x

\int \frac{7 - 6 x ^{3} + 2 x ^{6}}{x ^{6}} d x

\int tan^{4} (3 x) cos^{5} (3 x) d x

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

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

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