What is the integral of x(x-1)^4 ?

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

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integral of x(x-1)^4

\int x (x - 1)^{4} d x

\frac{x ^{6}}{6} - \frac{4 x ^{5}}{5} + \frac{3 x ^{4}}{2} - \frac{4 x ^{3}}{3} + \frac{x ^{2}}{2} + C

Solution steps

\int x (x - 1)^{4} d x

Expand

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

= \int x^{5} - 4 x^{4} + 6 x^{3} - 4 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^{5} d x - \int 4 x^{4} d x + \int 6 x^{3} d x - \int 4 x^{2} d x + \int x d x

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

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

\int 6 x^{3} d x = \frac{3 x ^{4}}{2}

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

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

= \frac{x ^{6}}{6} - \frac{4 x ^{5}}{5} + \frac{3 x ^{4}}{2} - \frac{4 x ^{3}}{3} + \frac{x ^{2}}{2}

Add a constant to the solution

= \frac{x ^{6}}{6} - \frac{4 x ^{5}}{5} + \frac{3 x ^{4}}{2} - \frac{4 x ^{3}}{3} + \frac{x ^{2}}{2} + C

f (x) = 4^{x^{9}}

\frac{d}{d x} (lo g_{e} (x))

\int \frac{9}{x ^{2} x ^{2} + 16} d x

tan^{2} (5 θ)

n = 1 \sum \infty 5 (- 1)^{n} n x^{n}

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