What is the integral of (6x+1)(x^2+x)^2 ?

The integral of (6x+1)(x^2+x)^2 is x^6+(13x^5)/5+2x^4+(x^3)/3+C

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integral of (6x+1)(x^2+x)^2

\int (6 x + 1) (x^{2} + x)^{2} d x

x^{6} + \frac{13 x ^{5}}{5} + 2 x^{4} + \frac{x ^{3}}{3} + C

Solution steps

\int (6 x + 1) (x^{2} + x)^{2} d x

Expand

(6 x + 1) (x^{2} + x)^{2} : 6 x^{5} + 13 x^{4} + 8 x^{3} + x^{2}

= \int 6 x^{5} + 13 x^{4} + 8 x^{3} + x^{2} 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 6 x^{5} d x + \int 13 x^{4} d x + \int 8 x^{3} d x + \int x^{2} d x

\int 6 x^{5} d x = x^{6}

\int 13 x^{4} d x = \frac{13 x ^{5}}{5}

\int 8 x^{3} d x = 2 x^{4}

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

= x^{6} + \frac{13 x ^{5}}{5} + 2 x^{4} + \frac{x ^{3}}{3}

Add a constant to the solution

= x^{6} + \frac{13 x ^{5}}{5} + 2 x^{4} + \frac{x ^{3}}{3} + C

What is the integral of (6x+1)(x^2+x)^2 ?
The integral of (6x+1)(x^2+x)^2 is x^6+(13x^5)/5+2x^4+(x^3)/3+C