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

The integral of 1/4 x^3+3x^2-3/4 x+4 is (x^4}{16}+x^3-\frac{3x^2)/8+4x+C

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

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

\frac{x ^{4}}{16} + x^{3} - \frac{3 x ^{2}}{8} + 4 x + C

Solution steps

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

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

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

\int \frac{3}{4} x d x = \frac{3 x ^{2}}{8}

\int 4 d x = 4 x

= \frac{x ^{4}}{16} + x^{3} - \frac{3 x ^{2}}{8} + 4 x

Add a constant to the solution

= \frac{x ^{4}}{16} + x^{3} - \frac{3 x ^{2}}{8} + 4 x + C

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What is the integral of 1/4 x^3+3x^2-3/4 x+4 ?
The integral of 1/4 x^3+3x^2-3/4 x+4 is (x^4}{16}+x^3-\frac{3x^2)/8+4x+C