What is the integral of-(3x^2}{32}+\frac{3x)/8 ?

The integral of-(3x^2}{32}+\frac{3x)/8 is -(x^3)/(32)+(3x^2)/(16)+C

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integral of-(3x^2}{32}+\frac{3x)/8

\int - \frac{3 x ^{2}}{32} + \frac{3 x}{8} d x

- \frac{x ^{3}}{32} + \frac{3 x ^{2}}{16} + C

Solution steps

\int - \frac{3 x ^{2}}{32} + \frac{3 x}{8} d x

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= - \int \frac{3 x ^{2}}{32} d x + \int \frac{3 x}{8} d x

\int \frac{3 x ^{2}}{32} d x = \frac{x ^{3}}{32}

\int \frac{3 x}{8} d x = \frac{3 x ^{2}}{16}

= - \frac{x ^{3}}{32} + \frac{3 x ^{2}}{16}

Add a constant to the solution

= - \frac{x ^{3}}{32} + \frac{3 x ^{2}}{16} + C

y = \frac{2 x + 1}{3 x - 4}

\frac{d}{d x} (e^{- \frac{x}{4}})

\int \frac{( x ^{2} )}{49 - x ^{2}} d x

y^{^{''}} - 9 y = 2 t^{2} + 3 t + 9

\frac{d}{d x} (3 x^{2} + e^{2 x} + π^{2})

What is the integral of-(3x^2}{32}+\frac{3x)/8 ?
The integral of-(3x^2}{32}+\frac{3x)/8 is -(x^3)/(32)+(3x^2)/(16)+C