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

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

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

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

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

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

Solution steps

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

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

\int 1 d x = x

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

Add a constant to the solution

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

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

\int \frac{( y + 3 )}{y ^{3} + 4 y ^{2}} d y

\int (x^{\frac{3}{2}} + 12 x + 2) d x

\int \frac{1}{1 - x + 3 1 - x} d x

\int \frac{6}{( x ^{2} + 1 ) ( x ^{2} + 4 )} d x

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