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

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

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

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

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

- x^{5} + \frac{6 x ^{5}}{5} - \frac{2 x ^{3}}{3} + x + C

Solution steps

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

Apply Integration By Parts

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

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

\int (4 x^{3} - 4 x) x d x = 2 x^{3} (x^{2} - 1) - 2 (\frac{3 x ^{5}}{5} - \frac{x ^{3}}{3})

= (x - 1)^{2} (x + 1)^{2} x - (2 x^{3} (x^{2} - 1) - 2 (\frac{3 x ^{5}}{5} - \frac{x ^{3}}{3}))

Simplify

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

= - x^{5} + \frac{6 x ^{5}}{5} - \frac{2 x ^{3}}{3} + x

Add a constant to the solution

= - x^{5} + \frac{6 x ^{5}}{5} - \frac{2 x ^{3}}{3} + x + C

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

\frac{4}{1 - x}

y = 8 x - 12, y = x^{2}, x = 1

x \to 0 lim (\frac{tan ( x )}{2 x})

t + 29

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