What is the integral of (2x^2-3x+2)^{3/2}(x-3/4) ?

The integral of (2x^2-3x+2)^{3/2}(x-3/4) is 1/10 (2x^2-3x+2)^{5/2}+C

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integral of (2x^2-3x+2)^{3/2}(x-3/4)

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

\frac{1}{10} (2 x^{2} - 3 x + 2)^{\frac{5}{2}} + C

Solution steps

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

Apply u-substitution

: \int \frac{u ^{\frac{3}{2}}}{4} d u

= \int \frac{u ^{\frac{3}{2}}}{4} d u

Take the constant out:

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

= \frac{1}{4} \cdot \int u^{\frac{3}{2}} d u

Apply the Power Rule

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

= \frac{1}{4} \cdot \frac{2}{5} u^{\frac{5}{2}}

Substitute back

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

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

Simplify

\frac{1}{4} \cdot \frac{2}{5} (2 x^{2} - 3 x + 2)^{\frac{5}{2}} : \frac{1}{10} (2 x^{2} - 3 x + 2)^{\frac{5}{2}}

= \frac{1}{10} (2 x^{2} - 3 x + 2)^{\frac{5}{2}}

Add a constant to the solution

= \frac{1}{10} (2 x^{2} - 3 x + 2)^{\frac{5}{2}} + C

\frac{\partial}{\partial y} (\frac{2 x - 2 z}{3 y + 2 z})

y = 2 x^{2} - 4 x + 2

\int \frac{( ln ( x ) ) ^{25}}{x} d x

\int_{1}^{2} w^{5} e^{w^{3}} d w

\frac{d y}{d x} - e^{2 x} = 3, y (0) = 4

What is the integral of (2x^2-3x+2)^{3/2}(x-3/4) ?
The integral of (2x^2-3x+2)^{3/2}(x-3/4) is 1/10 (2x^2-3x+2)^{5/2}+C