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

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

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integral of (3x^2-2x+1)(x^3-x^2+x)^4

\int (3 x^{2} - 2 x + 1) (x^{3} - x^{2} + x)^{4} d x

\frac{( x ^{3} - x ^{2} + x ) ^{5}}{5} + C

Solution steps

\int (3 x^{2} - 2 x + 1) (x^{3} - x^{2} + x)^{4} d x

Apply u-substitution

: \int u^{4} d u

= \int u^{4} d u

Apply the Power Rule

: \frac{u ^{5}}{5}

= \frac{u ^{5}}{5}

Substitute back

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

= \frac{( x ^{3} - x ^{2} + x ) ^{5}}{5}

Add a constant to the solution

= \frac{( x ^{3} - x ^{2} + x ) ^{5}}{5} + C

x \to 0 lim (\frac{1}{x} - \frac{1}{∣ x ∣})

x \to 3 lim (x^{2} + 7 x - 5)

\frac{d}{d x} (x^{2} cos (y))

\frac{d}{d x} ((x^{2} + 18) (2 x + 15))

y = x^{3} (9 - x^{2})

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