What is the integral of x(2x+1)^7 ?

The integral of x(2x+1)^7 is (128x^9)/9+56x^8+96x^7+(280x^6)/3+56x^5+21x^4+(14x^3)/3+(x^2)/2+C

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integral of x(2x+1)^7

\int x (2 x + 1)^{7} d x

\frac{128 x ^{9}}{9} + 56 x^{8} + 96 x^{7} + \frac{280 x ^{6}}{3} + 56 x^{5} + 21 x^{4} + \frac{14 x ^{3}}{3} + \frac{x ^{2}}{2} + C

Solution steps

\int x (2 x + 1)^{7} d x

Expand

x (2 x + 1)^{7} : 128 x^{8} + 448 x^{7} + 672 x^{6} + 560 x^{5} + 280 x^{4} + 84 x^{3} + 14 x^{2} + x

= \int 128 x^{8} + 448 x^{7} + 672 x^{6} + 560 x^{5} + 280 x^{4} + 84 x^{3} + 14 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 128 x^{8} d x + \int 448 x^{7} d x + \int 672 x^{6} d x + \int 560 x^{5} d x + \int 280 x^{4} d x + \int 84 x^{3} d x + \int 14 x^{2} d x + \int x d x

\int 128 x^{8} d x = \frac{128 x ^{9}}{9}

\int 448 x^{7} d x = 56 x^{8}

\int 672 x^{6} d x = 96 x^{7}

\int 560 x^{5} d x = \frac{280 x ^{6}}{3}

\int 280 x^{4} d x = 56 x^{5}

\int 84 x^{3} d x = 21 x^{4}

\int 14 x^{2} d x = \frac{14 x ^{3}}{3}

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

= \frac{128 x ^{9}}{9} + 56 x^{8} + 96 x^{7} + \frac{280 x ^{6}}{3} + 56 x^{5} + 21 x^{4} + \frac{14 x ^{3}}{3} + \frac{x ^{2}}{2}

Add a constant to the solution

= \frac{128 x ^{9}}{9} + 56 x^{8} + 96 x^{7} + \frac{280 x ^{6}}{3} + 56 x^{5} + 21 x^{4} + \frac{14 x ^{3}}{3} + \frac{x ^{2}}{2} + C

\frac{d y}{d x} = x (4 - y)

\int x^{2} + 2 x d x

\frac{d}{d x} (\frac{1}{( x + 4 ) ^{2}})

\frac{d}{d x} (sin (2) x)

n = 0 \sum \infty (ln (3))^{n}

What is the integral of x(2x+1)^7 ?
The integral of x(2x+1)^7 is (128x^9)/9+56x^8+96x^7+(280x^6)/3+56x^5+21x^4+(14x^3)/3+(x^2)/2+C