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

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

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integral of x(x-1)^6

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

\frac{1}{2} x^{2} (x - 1)^{6} - \frac{3 x ^{8}}{8} + \frac{15 x ^{7}}{7} - 5 x^{6} + 6 x^{5} - \frac{15 x ^{4}}{4} + x^{3} + C

Solution steps

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

Apply Integration By Parts

: \frac{1}{2} x^{2} (x - 1)^{6} - \int 3 x^{2} (x - 1)^{5} d x

= \frac{1}{2} x^{2} (x - 1)^{6} - \int 3 x^{2} (x - 1)^{5} d x

\int 3 x^{2} (x - 1)^{5} d x = \frac{3 x ^{8}}{8} - \frac{15 x ^{7}}{7} + 5 x^{6} - 6 x^{5} + \frac{15 x ^{4}}{4} - x^{3}

= \frac{1}{2} x^{2} (x - 1)^{6} - (\frac{3 x ^{8}}{8} - \frac{15 x ^{7}}{7} + 5 x^{6} - 6 x^{5} + \frac{15 x ^{4}}{4} - x^{3})

Simplify

= \frac{1}{2} x^{2} (x - 1)^{6} - \frac{3 x ^{8}}{8} + \frac{15 x ^{7}}{7} - 5 x^{6} + 6 x^{5} - \frac{15 x ^{4}}{4} + x^{3}

Add a constant to the solution

= \frac{1}{2} x^{2} (x - 1)^{6} - \frac{3 x ^{8}}{8} + \frac{15 x ^{7}}{7} - 5 x^{6} + 6 x^{5} - \frac{15 x ^{4}}{4} + x^{3} + C

x \to 1 + lim (\frac{∣ x - 1 ∣}{1 - x})

\int \frac{θ}{y ^{2}} e^{- \frac{θ}{y}} d y

\int \frac{5 x ^{2} - 5 x}{2 x ^{3} - 3 x ^{2} + 12} d x

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

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