What is the integral of (x-2)(x^2-4x+9)^8 ?

The integral of (x-2)(x^2-4x+9)^8 is 1/18 (x^2-4x+9)^9+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)(x^2-4x+9)^8

\int (x - 2) (x^{2} - 4 x + 9)^{8} d x

\frac{1}{18} (x^{2} - 4 x + 9)^{9} + C

Solution steps

\int (x - 2) (x^{2} - 4 x + 9)^{8} d x

Apply u-substitution

: \int u (u^{2} + 5)^{8} d u

= \int u (u^{2} + 5)^{8} d u

Apply u-substitution

: \int \frac{v ^{8}}{2} d v

= \int \frac{v ^{8}}{2} d v

Take the constant out:

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

= \frac{1}{2} \cdot \int v^{8} d v

Apply the Power Rule

: \frac{v ^{9}}{9}

= \frac{1}{2} \cdot \frac{v ^{9}}{9}

Substitute back

= \frac{1}{2} \cdot \frac{( ( x - 2 ) ^{2} + 5 ) ^{9}}{9}

Simplify

\frac{1}{2} \cdot \frac{( ( x - 2 ) ^{2} + 5 ) ^{9}}{9} : \frac{1}{18} (x^{2} - 4 x + 9)^{9}

= \frac{1}{18} (x^{2} - 4 x + 9)^{9}

Add a constant to the solution

= \frac{1}{18} (x^{2} - 4 x + 9)^{9} + C

\int 10 x + 3 d x

\int \frac{6 x}{( 9 + x ^{2} ) ^{3}} d x

\int 8 x + 5 d x

\int \frac{x}{2} 5 x + 1 d x

\int \frac{1}{4 ^{2} + x ^{2}} d x

What is the integral of (x-2)(x^2-4x+9)^8 ?
The integral of (x-2)(x^2-4x+9)^8 is 1/18 (x^2-4x+9)^9+C