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

The integral from 0 to 1 of x^3(1+x^4)^3 is 15/16

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integral from 0 to 1 of x^3(1+x^4)^3

\int_{0}^{1} x^{3} (1 + x^{4})^{3} d x

\frac{15}{16}

Decimal Notation

0.9375

Solution steps

\int_{0}^{1} x^{3} (1 + x^{4})^{3} d x

Expand

x^{3} (1 + x^{4})^{3} : x^{3} + 3 x^{7} + 3 x^{11} + x^{15}

= \int_{0}^{1} x^{3} + 3 x^{7} + 3 x^{11} + x^{15} 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^{3} d x + \int_{0}^{1} 3 x^{7} d x + \int_{0}^{1} 3 x^{11} d x + \int_{0}^{1} x^{15} d x

\int_{0}^{1} x^{3} d x = \frac{1}{4}

\int_{0}^{1} 3 x^{7} d x = \frac{3}{8}

\int_{0}^{1} 3 x^{11} d x = \frac{1}{4}

\int_{0}^{1} x^{15} d x = \frac{1}{16}

= \frac{1}{4} + \frac{3}{8} + \frac{1}{4} + \frac{1}{16}

Simplify

\frac{1}{4} + \frac{3}{8} + \frac{1}{4} + \frac{1}{16} : \frac{15}{16}

= \frac{15}{16}

ln (x + 2)

\int y^{2} e^{y^{3}} d y

\int x e^{\frac{x}{10}} d x

y = x^{18}

\frac{d}{d x} (x - 2 x + 2 ln (x + 1))

What is the integral from 0 to 1 of x^3(1+x^4)^3 ?
The integral from 0 to 1 of x^3(1+x^4)^3 is 15/16