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

The integral of (x^2-1)^4 is (x^9)/9-(4x^7)/7+(6x^5)/5-(4x^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^2-1)^4

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

\frac{x ^{9}}{9} - \frac{4 x ^{7}}{7} + \frac{6 x ^{5}}{5} - \frac{4 x ^{3}}{3} + x + C

Solution steps

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

Expand

(x^{2} - 1)^{4} : x^{8} - 4 x^{6} + 6 x^{4} - 4 x^{2} + 1

= \int x^{8} - 4 x^{6} + 6 x^{4} - 4 x^{2} + 1 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^{8} d x - \int 4 x^{6} d x + \int 6 x^{4} d x - \int 4 x^{2} d x + \int 1 d x

\int x^{8} d x = \frac{x ^{9}}{9}

\int 4 x^{6} d x = \frac{4 x ^{7}}{7}

\int 6 x^{4} d x = \frac{6 x ^{5}}{5}

\int 4 x^{2} d x = \frac{4 x ^{3}}{3}

\int 1 d x = x

= \frac{x ^{9}}{9} - \frac{4 x ^{7}}{7} + \frac{6 x ^{5}}{5} - \frac{4 x ^{3}}{3} + x

Add a constant to the solution

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

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

x^{12}

\int sin^{4} (5 x) cos^{6} (5 x) d x

\frac{d}{d x} (- 5 x^{- 6})

\int 11 tan^{3} (x) sec (x) d x

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