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

The integral of (1/2 x^3+3x^2-3/4 x+4) is (x^4)/8+x^3-(3x^2)/8+4x+C

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

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

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

Solution steps

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

\int \frac{1}{2} x^{3} d x = \frac{x ^{4}}{8}

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

Add a constant to the solution

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

\int (1 - 3 x)^{\frac{5}{2}} d x

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

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

\int \frac{sin ( x )}{3 + cos ( x )} d x

\int \frac{x}{2 x ^{2} + 8} d x

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