What is the integral of x^2(sqrt(x)+1) ?

The integral of x^2(sqrt(x)+1) is 2/7 x^{7/2}+(x^3)/3+C

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integral of x^2(sqrt(x)+1)

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

\frac{2}{7} x^{\frac{7}{2}} + \frac{x ^{3}}{3} + C

Solution steps

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

Expand

x^{2} (x + 1) : x^{\frac{5}{2}} + x^{2}

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

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

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

= \frac{2}{7} x^{\frac{7}{2}} + \frac{x ^{3}}{3}

Add a constant to the solution

= \frac{2}{7} x^{\frac{7}{2}} + \frac{x ^{3}}{3} + C

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What is the integral of x^2(sqrt(x)+1) ?
The integral of x^2(sqrt(x)+1) is 2/7 x^{7/2}+(x^3)/3+C