What is the integral of x(x+1)^6 ?

The integral of x(x+1)^6 is (x^8)/8+(6x^7)/7+(5x^6)/2+4x^5+(15x^4)/4+2x^3+(x^2)/2+C

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integral of x(x+1)^6

\int x (x + 1)^{6} d x

\frac{x ^{8}}{8} + \frac{6 x ^{7}}{7} + \frac{5 x ^{6}}{2} + 4 x^{5} + \frac{15 x ^{4}}{4} + 2 x^{3} + \frac{x ^{2}}{2} + C

Solution steps

\int x (x + 1)^{6} d x

Expand

x (x + 1)^{6} : x^{7} + 6 x^{6} + 15 x^{5} + 20 x^{4} + 15 x^{3} + 6 x^{2} + x

= \int x^{7} + 6 x^{6} + 15 x^{5} + 20 x^{4} + 15 x^{3} + 6 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 x^{7} d x + \int 6 x^{6} d x + \int 15 x^{5} d x + \int 20 x^{4} d x + \int 15 x^{3} d x + \int 6 x^{2} d x + \int x d x

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

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

\int 15 x^{5} d x = \frac{5 x ^{6}}{2}

\int 20 x^{4} d x = 4 x^{5}

\int 15 x^{3} d x = \frac{15 x ^{4}}{4}

\int 6 x^{2} d x = 2 x^{3}

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

= \frac{x ^{8}}{8} + \frac{6 x ^{7}}{7} + \frac{5 x ^{6}}{2} + 4 x^{5} + \frac{15 x ^{4}}{4} + 2 x^{3} + \frac{x ^{2}}{2}

Add a constant to the solution

= \frac{x ^{8}}{8} + \frac{6 x ^{7}}{7} + \frac{5 x ^{6}}{2} + 4 x^{5} + \frac{15 x ^{4}}{4} + 2 x^{3} + \frac{x ^{2}}{2} + C

\int (\frac{x}{x ^{2} - 2}) d x

10 cos (8 t) + 14 sin (8 t)

x \to 1 lim (2 x^{\frac{3}{2}} - 3 x^{2})

\frac{d}{d x} (7 x^{2} + 8 x)

What is the integral of x(x+1)^6 ?
The integral of x(x+1)^6 is (x^8)/8+(6x^7)/7+(5x^6)/2+4x^5+(15x^4)/4+2x^3+(x^2)/2+C