What is the integral of sqrt(x)(x^2-3x-5) ?

The integral of sqrt(x)(x^2-3x-5) is 2/7 x^{7/2}-6/5 x^{5/2}-10/3 x^{3/2}+C

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integral of sqrt(x)(x^2-3x-5)

\int x (x^{2} - 3 x - 5) d x

\frac{2}{7} x^{\frac{7}{2}} - \frac{6}{5} x^{\frac{5}{2}} - \frac{10}{3} x^{\frac{3}{2}} + C

Solution steps

\int x (x^{2} - 3 x - 5) d x

Expand

x (x^{2} - 3 x - 5) : x^{\frac{5}{2}} - 3 x^{\frac{3}{2}} - 5 x

= \int x^{\frac{5}{2}} - 3 x^{\frac{3}{2}} - 5 x 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^{\frac{5}{2}} d x - \int 3 x^{\frac{3}{2}} d x - \int 5 x d x

\int x^{\frac{5}{2}} d x = \frac{2}{7} x^{\frac{7}{2}}

\int 3 x^{\frac{3}{2}} d x = \frac{6}{5} x^{\frac{5}{2}}

\int 5 x d x = \frac{10}{3} x^{\frac{3}{2}}

= \frac{2}{7} x^{\frac{7}{2}} - \frac{6}{5} x^{\frac{5}{2}} - \frac{10}{3} x^{\frac{3}{2}}

Add a constant to the solution

= \frac{2}{7} x^{\frac{7}{2}} - \frac{6}{5} x^{\frac{5}{2}} - \frac{10}{3} x^{\frac{3}{2}} + C

f (x) = 2^{x}

\frac{\partial}{\partial s} (rs e^{t})

\int_{3}^{8} x^{3} - π x^{2} d x

g (x) = \frac{5}{x ^{3}}

\frac{d}{d x} ((2 x + 1)^{5} (3 x - 2)^{7})

What is the integral of sqrt(x)(x^2-3x-5) ?
The integral of sqrt(x)(x^2-3x-5) is 2/7 x^{7/2}-6/5 x^{5/2}-10/3 x^{3/2}+C