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

The integral of (2+x)^6 is 64x+96x^2+80x^3+40x^4+12x^5+2x^6+(x^7)/7+C

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integral of (2+x)^6

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

64 x + 96 x^{2} + 80 x^{3} + 40 x^{4} + 12 x^{5} + 2 x^{6} + \frac{x ^{7}}{7} + C

Solution steps

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

Expand

(2 + x)^{6} : 64 + 192 x + 240 x^{2} + 160 x^{3} + 60 x^{4} + 12 x^{5} + x^{6}

= \int 64 + 192 x + 240 x^{2} + 160 x^{3} + 60 x^{4} + 12 x^{5} + x^{6} 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 64 d x + \int 192 x d x + \int 240 x^{2} d x + \int 160 x^{3} d x + \int 60 x^{4} d x + \int 12 x^{5} d x + \int x^{6} d x

\int 64 d x = 64 x

\int 192 x d x = 96 x^{2}

\int 240 x^{2} d x = 80 x^{3}

\int 160 x^{3} d x = 40 x^{4}

\int 60 x^{4} d x = 12 x^{5}

\int 12 x^{5} d x = 2 x^{6}

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

= 64 x + 96 x^{2} + 80 x^{3} + 40 x^{4} + 12 x^{5} + 2 x^{6} + \frac{x ^{7}}{7}

Add a constant to the solution

= 64 x + 96 x^{2} + 80 x^{3} + 40 x^{4} + 12 x^{5} + 2 x^{6} + \frac{x ^{7}}{7} + C

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What is the integral of (2+x)^6 ?
The integral of (2+x)^6 is 64x+96x^2+80x^3+40x^4+12x^5+2x^6+(x^7)/7+C