What is the integral of x(6x^2+x^{1/2}) ?

The integral of x(6x^2+x^{1/2}) is (3x^4)/2+2/5 x^{5/2}+C

integral of x(6x^2+x^{1/2})

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

\frac{3 x ^{4}}{2} + \frac{2}{5} x^{\frac{5}{2}} + C

Solution steps

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

Expand

x (6 x^{2} + x^{\frac{1}{2}}) : 6 x^{3} + x^{\frac{3}{2}}

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

\int 6 x^{3} d x = \frac{3 x ^{4}}{2}

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

= \frac{3 x ^{4}}{2} + \frac{2}{5} x^{\frac{5}{2}}

Add a constant to the solution

= \frac{3 x ^{4}}{2} + \frac{2}{5} x^{\frac{5}{2}} + C

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

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

\int ln (x^{2} + 6 x + 10) d x

\int x e^{- 0.2 x^{2}} d x

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

What is the integral of x(6x^2+x^{1/2}) ?
The integral of x(6x^2+x^{1/2}) is (3x^4)/2+2/5 x^{5/2}+C