What is the integral of (x^2-3x+1)^{100}(2x-3) ?

The integral of (x^2-3x+1)^{100}(2x-3) is 1/(4^{100)*404}(4x^2-12x+4)^{101}+C

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integral of (x^2-3x+1)^{100}(2x-3)

\int (x^{2} - 3 x + 1)^{100} (2 x - 3) d x

\frac{1}{4 ^{100} \cdot 404} (4 x^{2} - 12 x + 4)^{101} + C

Solution steps

\int (x^{2} - 3 x + 1)^{100} (2 x - 3) d x

Apply u-substitution

= \int \frac{u ( u ^{2} - 5 ) ^{100}}{4 ^{100} \cdot 2} d u

Take the constant out:

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

= \frac{1}{4 ^{100} \cdot 2} \cdot \int u (u^{2} - 5)^{100} d u

Apply u-substitution

= \frac{1}{4 ^{100} \cdot 2} \cdot \int \frac{v ^{100}}{2} d v

Take the constant out:

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

= \frac{1}{4 ^{100} \cdot 2} \cdot \frac{1}{2} \cdot \int v^{100} d v

Apply the Power Rule

= \frac{1}{4 ^{100} \cdot 2} \cdot \frac{1}{2} \cdot \frac{v ^{101}}{101}

Substitute back

= \frac{1}{4 ^{100} \cdot 2} \cdot \frac{1}{2} \cdot \frac{( ( 2 x - 3 ) ^{2} - 5 ) ^{101}}{101}

Simplify

\frac{1}{4 ^{100} \cdot 2} \cdot \frac{1}{2} \cdot \frac{( ( 2 x - 3 ) ^{2} - 5 ) ^{101}}{101} : \frac{1}{4 ^{100} \cdot 404} (4 x^{2} - 12 x + 4)^{101}

= \frac{1}{4 ^{100} \cdot 404} (4 x^{2} - 12 x + 4)^{101}

Add a constant to the solution

= \frac{1}{4 ^{100} \cdot 404} (4 x^{2} - 12 x + 4)^{101} + C

What is the integral of (x^2-3x+1)^{100}(2x-3) ?
The integral of (x^2-3x+1)^{100}(2x-3) is 1/(4^{100)*404}(4x^2-12x+4)^{101}+C