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

The integral of x^2(1-x)^6 is -1/7 x^2(1-x)^7-1/252 (-8(1-x)^9+9(1-x)^8)+C

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integral of x^2(1-x)^6

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

- \frac{1}{7} x^{2} (1 - x)^{7} - \frac{1}{252} (- 8 (1 - x)^{9} + 9 (1 - x)^{8}) + C

Solution steps

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

Apply u-substitution

: \int - (- u + 1)^{2} u^{6} d u

= \int - (- u + 1)^{2} u^{6} d u

Take the constant out:

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

= - \int (- u + 1)^{2} u^{6} d u

Apply Integration By Parts

: \frac{1}{7} u^{7} (- u + 1)^{2} - \int - \frac{2}{7} u^{7} (- u + 1) d u

= - (\frac{1}{7} u^{7} (- u + 1)^{2} - \int - \frac{2}{7} u^{7} (- u + 1) d u)

\int - \frac{2}{7} u^{7} (- u + 1) d u = - \frac{1}{252} (- 8 u^{9} + 9 u^{8})

= - (\frac{1}{7} u^{7} (- u + 1)^{2} - (- \frac{1}{252} (- 8 u^{9} + 9 u^{8})))

Substitute back

u = 1 - x

= - (\frac{1}{7} (1 - x)^{7} (- (1 - x) + 1)^{2} - (- \frac{1}{252} (- 8 (1 - x)^{9} + 9 (1 - x)^{8})))

Simplify

- (\frac{1}{7} (1 - x)^{7} (- (1 - x) + 1)^{2} - (- \frac{1}{252} (- 8 (1 - x)^{9} + 9 (1 - x)^{8}))) : - \frac{1}{7} x^{2} (1 - x)^{7} - \frac{1}{252} (- 8 (1 - x)^{9} + 9 (1 - x)^{8})

= - \frac{1}{7} x^{2} (1 - x)^{7} - \frac{1}{252} (- 8 (1 - x)^{9} + 9 (1 - x)^{8})

Add a constant to the solution

= - \frac{1}{7} x^{2} (1 - x)^{7} - \frac{1}{252} (- 8 (1 - x)^{9} + 9 (1 - x)^{8}) + C

\frac{d}{d x} (- \frac{1}{x ^{7.7}})

\frac{d}{d x} (\frac{∣ x ∣}{x})

\int \frac{x ^{3} - 8 x + 1}{( 2 x ) ^{4}} d x

x y^{^{'}} + y = - 2 x y^{2}, y (1) = - 5

\frac{d}{d x} (tanh (x))

What is the integral of x^2(1-x)^6 ?
The integral of x^2(1-x)^6 is -1/7 x^2(1-x)^7-1/252 (-8(1-x)^9+9(1-x)^8)+C