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

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

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

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

\frac{3 x ^{2}}{4} + \frac{1}{x ^{2}} + C

Solution steps

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

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

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

Expand

\frac{3 x ^{4} - 4}{x ^{3}} : 3 x - \frac{4}{x ^{3}}

= \frac{1}{2} \cdot \int 3 x - \frac{4}{x ^{3}} 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

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

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

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

= \frac{1}{2} (\frac{3 x ^{2}}{2} - (- \frac{2}{x ^{2}}))

Simplify

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

= \frac{3 x ^{2}}{4} + \frac{1}{x ^{2}}

Add a constant to the solution

= \frac{3 x ^{2}}{4} + \frac{1}{x ^{2}} + C

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