What is the integral of (2x^3+x)(x^4+x^2) ?

The integral of (2x^3+x)(x^4+x^2) is (x^8)/4+(x^6)/2+(x^4)/4+C

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integral of (2x^3+x)(x^4+x^2)

\int (2 x^{3} + x) (x^{4} + x^{2}) d x

\frac{x ^{8}}{4} + \frac{x ^{6}}{2} + \frac{x ^{4}}{4} + C

Solution steps

\int (2 x^{3} + x) (x^{4} + x^{2}) d x

Expand

(2 x^{3} + x) (x^{4} + x^{2}) : 2 x^{7} + 3 x^{5} + x^{3}

= \int 2 x^{7} + 3 x^{5} + x^{3} 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 2 x^{7} d x + \int 3 x^{5} d x + \int x^{3} d x

\int 2 x^{7} d x = \frac{x ^{8}}{4}

\int 3 x^{5} d x = \frac{x ^{6}}{2}

\int x^{3} d x = \frac{x ^{4}}{4}

= \frac{x ^{8}}{4} + \frac{x ^{6}}{2} + \frac{x ^{4}}{4}

Add a constant to the solution

= \frac{x ^{8}}{4} + \frac{x ^{6}}{2} + \frac{x ^{4}}{4} + C

f (x) = - 2 x + 7

\int 7 x^{9} - 3 x^{6} + 10 x^{3} d x

\int \frac{1}{( x + 1 ) ^{5}} d x

x \to 0 lim (\frac{e ^{b x} - 1}{x})

x \to + 1 lim (\frac{x}{( x - 1 )})

What is the integral of (2x^3+x)(x^4+x^2) ?
The integral of (2x^3+x)(x^4+x^2) is (x^8)/4+(x^6)/2+(x^4)/4+C