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

The integral from 0 to 1 of x(x^2+1)^3 is 15/8

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

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

\frac{15}{8}

Decimal Notation

1.875

Solution steps

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

Expand

x (x^{2} + 1)^{3} : x^{7} + 3 x^{5} + 3 x^{3} + x

= \int_{0}^{1} x^{7} + 3 x^{5} + 3 x^{3} + 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_{0}^{1} x^{7} d x + \int_{0}^{1} 3 x^{5} d x + \int_{0}^{1} 3 x^{3} d x + \int_{0}^{1} x d x

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

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

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

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

= \frac{1}{8} + \frac{1}{2} + \frac{3}{4} + \frac{1}{2}

Simplify

\frac{1}{8} + \frac{1}{2} + \frac{3}{4} + \frac{1}{2} : \frac{15}{8}

= \frac{15}{8}

\int ln (e x) d x

x \to 1 lim (x^{2} - x + 1)

\int (x + 2)^{4} d x

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

\frac{5}{2 s ^{2} + 3 s + 1}

What is the integral from 0 to 1 of x(x^2+1)^3 ?
The integral from 0 to 1 of x(x^2+1)^3 is 15/8