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

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

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

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

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

Solution steps

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

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

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

\int 4 d x = 4 x

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

Add a constant to the solution

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

\int \frac{1}{( y ^{2} + a ^{2} ) ^{\frac{3}{2}}} d y

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\int 2 sin^{4} (x) d x

\int x^{2} + 3 x + 4 d x

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

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