What is the integral of 42x(x+1)^5 ?

The integral of 42x(x+1)^5 is 42((x^7)/7+(5x^6)/6+2x^5+(5x^4)/2+(5x^3)/3+(x^2)/2)+C

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integral of 42x(x+1)^5

\int 42 x (x + 1)^{5} d x

42 (\frac{x ^{7}}{7} + \frac{5 x ^{6}}{6} + 2 x^{5} + \frac{5 x ^{4}}{2} + \frac{5 x ^{3}}{3} + \frac{x ^{2}}{2}) + C

Solution steps

\int 42 x (x + 1)^{5} d x

Take the constant out:

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

= 42 \cdot \int x (x + 1)^{5} d x

Expand

x (x + 1)^{5} : x^{6} + 5 x^{5} + 10 x^{4} + 10 x^{3} + 5 x^{2} + x

= 42 \cdot \int x^{6} + 5 x^{5} + 10 x^{4} + 10 x^{3} + 5 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

= 42 (\int x^{6} d x + \int 5 x^{5} d x + \int 10 x^{4} d x + \int 10 x^{3} d x + \int 5 x^{2} d x + \int x d x)

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

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

\int 10 x^{4} d x = 2 x^{5}

\int 10 x^{3} d x = \frac{5 x ^{4}}{2}

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

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

= 42 (\frac{x ^{7}}{7} + \frac{5 x ^{6}}{6} + 2 x^{5} + \frac{5 x ^{4}}{2} + \frac{5 x ^{3}}{3} + \frac{x ^{2}}{2})

Add a constant to the solution

= 42 (\frac{x ^{7}}{7} + \frac{5 x ^{6}}{6} + 2 x^{5} + \frac{5 x ^{4}}{2} + \frac{5 x ^{3}}{3} + \frac{x ^{2}}{2}) + C

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What is the integral of 42x(x+1)^5 ?
The integral of 42x(x+1)^5 is 42((x^7)/7+(5x^6)/6+2x^5+(5x^4)/2+(5x^3)/3+(x^2)/2)+C