What is the integral of ((3x^4+5x-4))/(x^3d) ?

The integral of ((3x^4+5x-4))/(x^3d) is 1/d ((3x^2)/2-5/x+2/(x^2))+C

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integral of ((3x^4+5x-4))/(x^3d)

integral

\frac{( 3 x ^{4} + 5 x - 4 )}{x ^{3} d}

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

Solution steps

\int \frac{3 x ^{4} + 5 x - 4}{x ^{3} d} d x

Take the constant out:

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

= \frac{1}{d} \cdot \int \frac{3 x ^{4} + 5 x - 4}{x ^{3}} d x

Expand

\frac{3 x ^{4} + 5 x - 4}{x ^{3}} : 3 x + \frac{5}{x ^{2}} - \frac{4}{x ^{3}}

= \frac{1}{d} \cdot \int 3 x + \frac{5}{x ^{2}} - \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}{d} (\int 3 x d x + \int \frac{5}{x ^{2}} d x - \int \frac{4}{x ^{3}} d x)

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

\int \frac{5}{x ^{2}} d x = - \frac{5}{x}

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

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

Simplify

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

Add a constant to the solution

= \frac{1}{d} (\frac{3 x ^{2}}{2} - \frac{5}{x} + \frac{2}{x ^{2}}) + C

cos (\frac{2}{t})

arctan (x^{2})

\frac{( e ^{- x} ) ^{2}}{b}

5 t^{2} - t

\frac{1}{ln ^{2} ( x )}

What is the integral of ((3x^4+5x-4))/(x^3d) ?
The integral of ((3x^4+5x-4))/(x^3d) is 1/d ((3x^2)/2-5/x+2/(x^2))+C