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

The integral of 2x(2x+1)^7 is (256x^9)/9+112x^8+192x^7+(560x^6)/3+112x^5+42x^4+(28x^3)/3+x^2+C

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integral of 2x(2x+1)^7

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

\frac{256 x ^{9}}{9} + 112 x^{8} + 192 x^{7} + \frac{560 x ^{6}}{3} + 112 x^{5} + 42 x^{4} + \frac{28 x ^{3}}{3} + x^{2} + C

Solution steps

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

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= 2 \cdot \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

= 2 \cdot \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

= 2 (\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}

= 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})

Simplify

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}) : \frac{256 x ^{9}}{9} + 112 x^{8} + 192 x^{7} + \frac{560 x ^{6}}{3} + 112 x^{5} + 42 x^{4} + \frac{28 x ^{3}}{3} + x^{2}

= \frac{256 x ^{9}}{9} + 112 x^{8} + 192 x^{7} + \frac{560 x ^{6}}{3} + 112 x^{5} + 42 x^{4} + \frac{28 x ^{3}}{3} + x^{2}

Add a constant to the solution

= \frac{256 x ^{9}}{9} + 112 x^{8} + 192 x^{7} + \frac{560 x ^{6}}{3} + 112 x^{5} + 42 x^{4} + \frac{28 x ^{3}}{3} + x^{2} + C

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What is the integral of 2x(2x+1)^7 ?
The integral of 2x(2x+1)^7 is (256x^9)/9+112x^8+192x^7+(560x^6)/3+112x^5+42x^4+(28x^3)/3+x^2+C