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

The integral of x(x+2)^6 is (x^8)/8+(12x^7)/7+10x^6+32x^5+60x^4+64x^3+32x^2+C

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integral of x(x+2)^6

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

\frac{x ^{8}}{8} + \frac{12 x ^{7}}{7} + 10 x^{6} + 32 x^{5} + 60 x^{4} + 64 x^{3} + 32 x^{2} + C

Solution steps

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

Expand

x (x + 2)^{6} : x^{7} + 12 x^{6} + 60 x^{5} + 160 x^{4} + 240 x^{3} + 192 x^{2} + 64 x

= \int x^{7} + 12 x^{6} + 60 x^{5} + 160 x^{4} + 240 x^{3} + 192 x^{2} + 64 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 x^{7} d x + \int 12 x^{6} d x + \int 60 x^{5} d x + \int 160 x^{4} d x + \int 240 x^{3} d x + \int 192 x^{2} d x + \int 64 x d x

\int x^{7} d x = \frac{x ^{8}}{8}

\int 12 x^{6} d x = \frac{12 x ^{7}}{7}

\int 60 x^{5} d x = 10 x^{6}

\int 160 x^{4} d x = 32 x^{5}

\int 240 x^{3} d x = 60 x^{4}

\int 192 x^{2} d x = 64 x^{3}

\int 64 x d x = 32 x^{2}

= \frac{x ^{8}}{8} + \frac{12 x ^{7}}{7} + 10 x^{6} + 32 x^{5} + 60 x^{4} + 64 x^{3} + 32 x^{2}

Add a constant to the solution

= \frac{x ^{8}}{8} + \frac{12 x ^{7}}{7} + 10 x^{6} + 32 x^{5} + 60 x^{4} + 64 x^{3} + 32 x^{2} + C

\int_{- x}^{x^{2}} e^{t^{2}} d t

\int 7 x + 2 d x

\frac{5 s e ^{- s}}{2 s ^{2} + 32}

\int \frac{x + 1}{( x + 2 ) ^{3}} d x

9 y^{^{''}} - 12 y^{^{'}} + 4 y = e^{- 3 x}

What is the integral of x(x+2)^6 ?
The integral of x(x+2)^6 is (x^8)/8+(12x^7)/7+10x^6+32x^5+60x^4+64x^3+32x^2+C