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

The integral of-1/4 x^3+3/2 x^2+1 is -(x^4}{16}+\frac{x^3)/2+x+C

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integral of-1/4 x^3+3/2 x^2+1

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

- \frac{x ^{4}}{16} + \frac{x ^{3}}{2} + x + C

Solution steps

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

\int \frac{1}{4} x^{3} d x = \frac{x ^{4}}{16}

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

\int 1 d x = x

= - \frac{x ^{4}}{16} + \frac{x ^{3}}{2} + x

Add a constant to the solution

= - \frac{x ^{4}}{16} + \frac{x ^{3}}{2} + x + C

What is the integral of-1/4 x^3+3/2 x^2+1 ?
The integral of-1/4 x^3+3/2 x^2+1 is -(x^4}{16}+\frac{x^3)/2+x+C