What is the integral of 4x(1-x)^6 ?

The integral of 4x(1-x)^6 is 2x^2(-1+x)^6-(3x^8)/2+(60x^7)/7-20x^6+24x^5-15x^4+4x^3+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral of 4x(1-x)^6

\int 4 x (1 - x)^{6} d x

2 x^{2} (- 1 + x)^{6} - \frac{3 x ^{8}}{2} + \frac{60 x ^{7}}{7} - 20 x^{6} + 24 x^{5} - 15 x^{4} + 4 x^{3} + C

Solution steps

\int 4 x (1 - x)^{6} d x

Take the constant out:

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

= 4 \cdot \int x (1 - x)^{6} d x

Factor

1 - x : - (- 1 + x)

= 4 \cdot \int x (- (- 1 + x))^{6} d x

(- (- 1 + x))^{6} = (- 1 + x)^{6}

= 4 \cdot \int x (- 1 + x)^{6} d x

Apply Integration By Parts

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

= 4 (\frac{1}{2} x^{2} (- 1 + x)^{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}

= 4 (\frac{1}{2} x^{2} (- 1 + x)^{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

4 (\frac{1}{2} x^{2} (- 1 + x)^{6} - (\frac{3 x ^{8}}{8} - \frac{15 x ^{7}}{7} + 5 x^{6} - 6 x^{5} + \frac{15 x ^{4}}{4} - x^{3})) : 2 x^{2} (- 1 + x)^{6} - \frac{3 x ^{8}}{2} + \frac{60 x ^{7}}{7} - 20 x^{6} + 24 x^{5} - 15 x^{4} + 4 x^{3}

= 2 x^{2} (- 1 + x)^{6} - \frac{3 x ^{8}}{2} + \frac{60 x ^{7}}{7} - 20 x^{6} + 24 x^{5} - 15 x^{4} + 4 x^{3}

Add a constant to the solution

= 2 x^{2} (- 1 + x)^{6} - \frac{3 x ^{8}}{2} + \frac{60 x ^{7}}{7} - 20 x^{6} + 24 x^{5} - 15 x^{4} + 4 x^{3} + C

\int \frac{2 x - 1}{x ( x - 1 )} d x

\int \frac{x}{( x ^{2} + 5 ) ^{3}} d x

\int x 3 x^{2} - 1 d x

\int ln (5 x + 6) d x

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

What is the integral of 4x(1-x)^6 ?
The integral of 4x(1-x)^6 is 2x^2(-1+x)^6-(3x^8)/2+(60x^7)/7-20x^6+24x^5-15x^4+4x^3+C