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

The integral of (3x^2-2x+1)(x^3-x^2+x)^3 is ((x^3-x^2+x)^4)/4+C

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

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

\frac{( x ^{3} - x ^{2} + x ) ^{4}}{4} + C

Solution steps

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

Apply u-substitution

: \int u^{3} d u

= \int u^{3} d u

Apply the Power Rule

: \frac{u ^{4}}{4}

= \frac{u ^{4}}{4}

Substitute back

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

= \frac{( x ^{3} - x ^{2} + x ) ^{4}}{4}

Add a constant to the solution

= \frac{( x ^{3} - x ^{2} + x ) ^{4}}{4} + C

y = \frac{9 + x f ( x )}{x}

\frac{d}{d x} (3 x^{2} sin (x))

\frac{d}{d x} (\frac{x}{x ^{2} + 2})

\frac{1}{x} sin (x)

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