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

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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

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