What is the integral of (x^2-3x+5)^{3/2}(2x-3) ?

The integral of (x^2-3x+5)^{3/2}(2x-3) is 2/5 (x^2-3x+5)^{5/2}+C

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integral of (x^2-3x+5)^{3/2}(2x-3)

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

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

Solution steps

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

Apply u-substitution

: \int \frac{u ( u ^{2} + 11 ) ^{\frac{3}{2}}}{16} d u

= \int \frac{u ( u ^{2} + 11 ) ^{\frac{3}{2}}}{16} d u

Take the constant out:

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

= \frac{1}{16} \cdot \int u (u^{2} + 11)^{\frac{3}{2}} d u

Apply u-substitution

: \int \frac{v ^{\frac{3}{2}}}{2} d v

= \frac{1}{16} \cdot \int \frac{v ^{\frac{3}{2}}}{2} d v

Take the constant out:

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

= \frac{1}{16} \cdot \frac{1}{2} \cdot \int v^{\frac{3}{2}} d v

Apply the Power Rule

: \frac{2}{5} v^{\frac{5}{2}}

= \frac{1}{16} \cdot \frac{1}{2} \cdot \frac{2}{5} v^{\frac{5}{2}}

Substitute back

= \frac{1}{16} \cdot \frac{1}{2} \cdot \frac{2}{5} ((2 x - 3)^{2} + 11)^{\frac{5}{2}}

Simplify

\frac{1}{16} \cdot \frac{1}{2} \cdot \frac{2}{5} ((2 x - 3)^{2} + 11)^{\frac{5}{2}} : \frac{2}{5} (x^{2} - 3 x + 5)^{\frac{5}{2}}

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

Add a constant to the solution

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

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