What is the integral of (6x^3-3x^2+5+7x^{-1})/(x^3) ?

The integral of (6x^3-3x^2+5+7x^{-1})/(x^3) is 6x-3ln|x|-5/(2x^2)-7/(3x^3)+C

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integral of (6x^3-3x^2+5+7x^{-1})/(x^3)

\int \frac{6 x ^{3} - 3 x ^{2} + 5 + 7 x ^{- 1}}{x ^{3}} d x

6 x - 3 ln ∣ x ∣ - \frac{5}{2 x ^{2}} - \frac{7}{3 x ^{3}} + C

Solution steps

\int \frac{6 x ^{3} - 3 x ^{2} + 5 + 7 x ^{- 1}}{x ^{3}} d x

Expand

\frac{6 x ^{3} - 3 x ^{2} + 5 + 7 x ^{- 1}}{x ^{3}} : 6 - \frac{3}{x} + \frac{5}{x ^{3}} + \frac{7}{x ^{4}}

= \int 6 - \frac{3}{x} + \frac{5}{x ^{3}} + \frac{7}{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 6 d x - \int \frac{3}{x} d x + \int \frac{5}{x ^{3}} d x + \int \frac{7}{x ^{4}} d x

\int 6 d x = 6 x

\int \frac{3}{x} d x = 3 ln ∣ x ∣

\int \frac{5}{x ^{3}} d x = - \frac{5}{2 x ^{2}}

\int \frac{7}{x ^{4}} d x = - \frac{7}{3 x ^{3}}

= 6 x - 3 ln ∣ x ∣ - \frac{5}{2 x ^{2}} - \frac{7}{3 x ^{3}}

Add a constant to the solution

= 6 x - 3 ln ∣ x ∣ - \frac{5}{2 x ^{2}} - \frac{7}{3 x ^{3}} + C

\int \frac{c}{x} e^{- 9 x} d x

\int \frac{1 + x}{x} d x

f (x) = x^{5 x}

y = x^{4} + 9 e^{x}, (0, 9)

2 y d x + (y - 2 x) d y = 0

What is the integral of (6x^3-3x^2+5+7x^{-1})/(x^3) ?
The integral of (6x^3-3x^2+5+7x^{-1})/(x^3) is 6x-3ln|x|-5/(2x^2)-7/(3x^3)+C