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

The integral of (3x^2-2x+1)(x^3-x^2+x)^9 is ((x^3-x^2+x)^{10})/(10)+C

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integral of (3x^2-2x+1)(x^3-x^2+x)^9

\int (3 x^{2} - 2 x + 1) (x^{3} - x^{2} + x)^{9} d x

\frac{( x ^{3} - x ^{2} + x ) ^{10}}{10} + C

Solution steps

\int (3 x^{2} - 2 x + 1) (x^{3} - x^{2} + x)^{9} d x

Apply u-substitution

: \int u^{9} d u

= \int u^{9} d u

Apply the Power Rule

: \frac{u ^{10}}{10}

= \frac{u ^{10}}{10}

Substitute back

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

= \frac{( x ^{3} - x ^{2} + x ) ^{10}}{10}

Add a constant to the solution

= \frac{( x ^{3} - x ^{2} + x ) ^{10}}{10} + C

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What is the integral of (3x^2-2x+1)(x^3-x^2+x)^9 ?
The integral of (3x^2-2x+1)(x^3-x^2+x)^9 is ((x^3-x^2+x)^{10})/(10)+C