What is the integral of (3x^2-2x+1)(x^3-x^2+x)^8 ?

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

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integral of (3x^2-2x+1)(x^3-x^2+x)^8

\int (3 x^{2} - 2 x + 1) (x^{3} - x^{2} + x)^{8} d x

\frac{( x ^{3} - x ^{2} + x ) ^{9}}{9} + C

Solution steps

\int (3 x^{2} - 2 x + 1) (x^{3} - x^{2} + x)^{8} d x

Apply u-substitution

: \int u^{8} d u

= \int u^{8} d u

Apply the Power Rule

: \frac{u ^{9}}{9}

= \frac{u ^{9}}{9}

Substitute back

u = x^{3} - x^{2} + x

= \frac{( x ^{3} - x ^{2} + x ) ^{9}}{9}

Add a constant to the solution

= \frac{( x ^{3} - x ^{2} + x ) ^{9}}{9} + C

y = (t \cdot sin (t))^{2} + t \cdot sin (t)

sec^{2} (2 x) - tan^{2} (2 x)

\frac{\partial}{\partial x} (- 4 e^{- x - y})

3 \frac{d y}{d x} + 12 y = 4

x \to 3 lim (\frac{x}{x - 2})

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