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

The integral from 1 to 2 of x(1+x^2)^3 is 609/8

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

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

\frac{609}{8}

Decimal Notation

76.125

Solution steps

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

Expand

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

= \int_{1}^{2} x + 3 x^{3} + 3 x^{5} + x^{7} 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_{1}^{2} x d x + \int_{1}^{2} 3 x^{3} d x + \int_{1}^{2} 3 x^{5} d x + \int_{1}^{2} x^{7} d x

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

\int_{1}^{2} 3 x^{3} d x = \frac{45}{4}

\int_{1}^{2} 3 x^{5} d x = \frac{63}{2}

\int_{1}^{2} x^{7} d x = \frac{255}{8}

= \frac{3}{2} + \frac{45}{4} + \frac{63}{2} + \frac{255}{8}

Simplify

\frac{3}{2} + \frac{45}{4} + \frac{63}{2} + \frac{255}{8} : \frac{609}{8}

= \frac{609}{8}

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What is the integral from 1 to 2 of x(1+x^2)^3 ?
The integral from 1 to 2 of x(1+x^2)^3 is 609/8