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

The integral from 1 to 2 of x(x^2+1)^2 is 39/2

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

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

\frac{39}{2}

Decimal Notation

19.5

Solution steps

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

Expand

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

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

\int_{1}^{2} x^{5} d x = \frac{21}{2}

\int_{1}^{2} 2 x^{3} d x = \frac{15}{2}

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

= \frac{21}{2} + \frac{15}{2} + \frac{3}{2}

Simplify

\frac{21}{2} + \frac{15}{2} + \frac{3}{2} : \frac{39}{2}

= \frac{39}{2}

\frac{d v}{d t} = \frac{( 10 - v )}{1}

\int \frac{1}{x ( x - 1 ) ^{2}} d x

\int 16 - y d y

\int_{0}^{2} - 10 x^{3} x^{2} + 4 d x

6 t \frac{d y}{d t} + y = t^{7}

What is the integral from 1 to 2 of x(x^2+1)^2 ?
The integral from 1 to 2 of x(x^2+1)^2 is 39/2