What is the integral of x(2x+1)^5 ?

The integral of x(2x+1)^5 is (32x^7)/7+(40x^6)/3+16x^5+10x^4+(10x^3)/3+(x^2)/2+C

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integral of x(2x+1)^5

\int x (2 x + 1)^{5} d x

\frac{32 x ^{7}}{7} + \frac{40 x ^{6}}{3} + 16 x^{5} + 10 x^{4} + \frac{10 x ^{3}}{3} + \frac{x ^{2}}{2} + C

Solution steps

\int x (2 x + 1)^{5} d x

Expand

x (2 x + 1)^{5} : 32 x^{6} + 80 x^{5} + 80 x^{4} + 40 x^{3} + 10 x^{2} + x

= \int 32 x^{6} + 80 x^{5} + 80 x^{4} + 40 x^{3} + 10 x^{2} + 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 32 x^{6} d x + \int 80 x^{5} d x + \int 80 x^{4} d x + \int 40 x^{3} d x + \int 10 x^{2} d x + \int x d x

\int 32 x^{6} d x = \frac{32 x ^{7}}{7}

\int 80 x^{5} d x = \frac{40 x ^{6}}{3}

\int 80 x^{4} d x = 16 x^{5}

\int 40 x^{3} d x = 10 x^{4}

\int 10 x^{2} d x = \frac{10 x ^{3}}{3}

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

= \frac{32 x ^{7}}{7} + \frac{40 x ^{6}}{3} + 16 x^{5} + 10 x^{4} + \frac{10 x ^{3}}{3} + \frac{x ^{2}}{2}

Add a constant to the solution

= \frac{32 x ^{7}}{7} + \frac{40 x ^{6}}{3} + 16 x^{5} + 10 x^{4} + \frac{10 x ^{3}}{3} + \frac{x ^{2}}{2} + C

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What is the integral of x(2x+1)^5 ?
The integral of x(2x+1)^5 is (32x^7)/7+(40x^6)/3+16x^5+10x^4+(10x^3)/3+(x^2)/2+C