What is the integral from 0 to 1 of x^2(x^3+1)^3 ?

The integral from 0 to 1 of x^2(x^3+1)^3 is 5/4

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integral from 0 to 1 of x^2(x^3+1)^3

\int_{0}^{1} x^{2} (x^{3} + 1)^{3} d x

\frac{5}{4}

Decimal

1.25

Solution steps

\int_{0}^{1} x^{2} (x^{3} + 1)^{3} d x

Expand

x^{2} (x^{3} + 1)^{3} : x^{11} + 3 x^{8} + 3 x^{5} + x^{2}

= \int_{0}^{1} x^{11} + 3 x^{8} + 3 x^{5} + 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_{0}^{1} x^{11} d x + \int_{0}^{1} 3 x^{8} d x + \int_{0}^{1} 3 x^{5} d x + \int_{0}^{1} x^{2} d x

\int_{0}^{1} x^{11} d x = \frac{1}{12}

\int_{0}^{1} 3 x^{8} d x = \frac{1}{3}

\int_{0}^{1} 3 x^{5} d x = \frac{1}{2}

\int_{0}^{1} x^{2} d x = \frac{1}{3}

= \frac{1}{12} + \frac{1}{3} + \frac{1}{2} + \frac{1}{3}

Simplify

\frac{1}{12} + \frac{1}{3} + \frac{1}{2} + \frac{1}{3} : \frac{5}{4}

= \frac{5}{4}

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

1000 \cdot \frac{1}{s ^{2} + 1000 s}

\int \frac{x ^{4}}{4} - \frac{x ^{3}}{3} - \frac{x ^{2}}{2} + x + C d x

\int 2 x^{3} d x

\frac{3}{ln ( x )}

What is the integral from 0 to 1 of x^2(x^3+1)^3 ?
The integral from 0 to 1 of x^2(x^3+1)^3 is 5/4