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

The integral of x(x-1)^5 is ((x-1)^7)/7+((x-1)^6)/6+C

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integral of x(x-1)^5

\int x (x - 1)^{5} d x

\frac{( x - 1 ) ^{7}}{7} + \frac{( x - 1 ) ^{6}}{6} + C

Solution steps

\int x (x - 1)^{5} d x

Apply u-substitution

: \int (u + 1) u^{5} d u

= \int (u + 1) u^{5} d u

Expand

(u + 1) u^{5} : u^{6} + u^{5}

= \int u^{6} + u^{5} d u

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= \int u^{6} d u + \int u^{5} d u

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

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

= \frac{u ^{7}}{7} + \frac{u ^{6}}{6}

Substitute back

u = x - 1

= \frac{( x - 1 ) ^{7}}{7} + \frac{( x - 1 ) ^{6}}{6}

Add a constant to the solution

= \frac{( x - 1 ) ^{7}}{7} + \frac{( x - 1 ) ^{6}}{6} + C

\frac{\partial}{\partial x} (y^{\frac{1}{3}} x^{\frac{2}{3}})

\int \frac{1}{( x - 1 ) ( x + 1 )} d x

f (x) = x^{3} - 2 x^{2} - 4 x - 6

\int (2 x + 5) e^{2 x + 5} d x

- \frac{9}{2} x^{2} + 3 x + 3

What is the integral of x(x-1)^5 ?
The integral of x(x-1)^5 is ((x-1)^7)/7+((x-1)^6)/6+C