What is the integral of x(x-3)^2(x+1)^4 ?

The integral of x(x-3)^2(x+1)^4 is (x^8)/8-(2x^7)/7-(3x^6)/2+(4x^5)/5+(31x^4)/4+10x^3+(9x^2)/2+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

integral of x(x-3)^2(x+1)^4

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

\frac{x ^{8}}{8} - \frac{2 x ^{7}}{7} - \frac{3 x ^{6}}{2} + \frac{4 x ^{5}}{5} + \frac{31 x ^{4}}{4} + 10 x^{3} + \frac{9 x ^{2}}{2} + C

Solution steps

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

Expand

x (x - 3)^{2} (x + 1)^{4} : x^{7} - 2 x^{6} - 9 x^{5} + 4 x^{4} + 31 x^{3} + 30 x^{2} + 9 x

= \int x^{7} - 2 x^{6} - 9 x^{5} + 4 x^{4} + 31 x^{3} + 30 x^{2} + 9 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^{7} d x - \int 2 x^{6} d x - \int 9 x^{5} d x + \int 4 x^{4} d x + \int 31 x^{3} d x + \int 30 x^{2} d x + \int 9 x d x

\int x^{7} d x = \frac{x ^{8}}{8}

\int 2 x^{6} d x = \frac{2 x ^{7}}{7}

\int 9 x^{5} d x = \frac{3 x ^{6}}{2}

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

\int 31 x^{3} d x = \frac{31 x ^{4}}{4}

\int 30 x^{2} d x = 10 x^{3}

\int 9 x d x = \frac{9 x ^{2}}{2}

= \frac{x ^{8}}{8} - \frac{2 x ^{7}}{7} - \frac{3 x ^{6}}{2} + \frac{4 x ^{5}}{5} + \frac{31 x ^{4}}{4} + 10 x^{3} + \frac{9 x ^{2}}{2}

Add a constant to the solution

= \frac{x ^{8}}{8} - \frac{2 x ^{7}}{7} - \frac{3 x ^{6}}{2} + \frac{4 x ^{5}}{5} + \frac{31 x ^{4}}{4} + 10 x^{3} + \frac{9 x ^{2}}{2} + C

What is the integral of x(x-3)^2(x+1)^4 ?
The integral of x(x-3)^2(x+1)^4 is (x^8)/8-(2x^7)/7-(3x^6)/2+(4x^5)/5+(31x^4)/4+10x^3+(9x^2)/2+C