What is the integral of (x+1/2)^6 ?

The integral of (x+1/2)^6 is (x^7)/7+(x^6)/2+(3x^5)/4+(5x^4)/8+(5x^3)/(16)+(3x^2)/(32)+1/64 x+C

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integral of (x+1/2)^6

\int (x + \frac{1}{2})^{6} d x

\frac{x ^{7}}{7} + \frac{x ^{6}}{2} + \frac{3 x ^{5}}{4} + \frac{5 x ^{4}}{8} + \frac{5 x ^{3}}{16} + \frac{3 x ^{2}}{32} + \frac{1}{64} x + C

Solution steps

\int (x + \frac{1}{2})^{6} d x

Expand

(x + \frac{1}{2})^{6} : x^{6} + 3 x^{5} + \frac{15 x ^{4}}{4} + \frac{5 x ^{3}}{2} + \frac{15 x ^{2}}{16} + \frac{3 x}{16} + \frac{1}{64}

= \int x^{6} + 3 x^{5} + \frac{15 x ^{4}}{4} + \frac{5 x ^{3}}{2} + \frac{15 x ^{2}}{16} + \frac{3 x}{16} + \frac{1}{64} 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^{6} d x + \int 3 x^{5} d x + \int \frac{15 x ^{4}}{4} d x + \int \frac{5 x ^{3}}{2} d x + \int \frac{15 x ^{2}}{16} d x + \int \frac{3 x}{16} d x + \int \frac{1}{64} d x

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

\int 3 x^{5} d x = \frac{x ^{6}}{2}

\int \frac{15 x ^{4}}{4} d x = \frac{3 x ^{5}}{4}

\int \frac{5 x ^{3}}{2} d x = \frac{5 x ^{4}}{8}

\int \frac{15 x ^{2}}{16} d x = \frac{5 x ^{3}}{16}

\int \frac{3 x}{16} d x = \frac{3 x ^{2}}{32}

\int \frac{1}{64} d x = \frac{1}{64} x

= \frac{x ^{7}}{7} + \frac{x ^{6}}{2} + \frac{3 x ^{5}}{4} + \frac{5 x ^{4}}{8} + \frac{5 x ^{3}}{16} + \frac{3 x ^{2}}{32} + \frac{1}{64} x

Add a constant to the solution

= \frac{x ^{7}}{7} + \frac{x ^{6}}{2} + \frac{3 x ^{5}}{4} + \frac{5 x ^{4}}{8} + \frac{5 x ^{3}}{16} + \frac{3 x ^{2}}{32} + \frac{1}{64} x + C

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What is the integral of (x+1/2)^6 ?
The integral of (x+1/2)^6 is (x^7)/7+(x^6)/2+(3x^5)/4+(5x^4)/8+(5x^3)/(16)+(3x^2)/(32)+1/64 x+C