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

The integral of sqrt(x)sqrt(x^{3/2)}-1 is 4/9 x^{9/4}-x+C

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

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

\frac{4}{9} x^{\frac{9}{4}} - x + C

Solution steps

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

x^{\frac{3}{2}} = x x^{0.5}

= \int x x x^{0.5} - 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 x x x^{0.5} d x - \int 1 d x

\int x x x^{0.5} d x = \frac{4}{9} x^{\frac{9}{4}}

\int 1 d x = x

= \frac{4}{9} x^{\frac{9}{4}} - x

Add a constant to the solution

= \frac{4}{9} x^{\frac{9}{4}} - x + C

sin (x), cos (x), x = - \frac{π}{6}

\frac{d}{d x} (x^{2} e^{5 x})

7 + 4 x

\int_{0}^{4} - x^{2} d x

x \to 2 lim (4 - 2 x)

What is the integral of sqrt(x)sqrt(x^{3/2)}-1 ?
The integral of sqrt(x)sqrt(x^{3/2)}-1 is 4/9 x^{9/4}-x+C