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

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

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

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

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

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

Solution steps

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

Expand

(3 x^{2} - 2 x + 1) (x^{3} - x^{2} + x)^{2} : 3 x^{8} - 8 x^{7} + 14 x^{6} - 14 x^{5} + 10 x^{4} - 4 x^{3} + x^{2}

= \int 3 x^{8} - 8 x^{7} + 14 x^{6} - 14 x^{5} + 10 x^{4} - 4 x^{3} + x^{2} 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 3 x^{8} d x - \int 8 x^{7} d x + \int 14 x^{6} d x - \int 14 x^{5} d x + \int 10 x^{4} d x - \int 4 x^{3} d x + \int x^{2} d x

\int 3 x^{8} d x = \frac{x ^{9}}{3}

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

\int 14 x^{6} d x = 2 x^{7}

\int 14 x^{5} d x = \frac{7 x ^{6}}{3}

\int 10 x^{4} d x = 2 x^{5}

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

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

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

Add a constant to the solution

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

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

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

\int \frac{5 e ^{- 3 + l n (3 x)}}{x} d x

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

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

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