What is the integral of (x^6-x)/(x^3) ?

The integral of (x^6-x)/(x^3) is (x^4)/4+1/x+C

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integral of (x^6-x)/(x^3)

\int \frac{x ^{6} - x}{x ^{3}} d x

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

Solution steps

\int \frac{x ^{6} - x}{x ^{3}} d x

Expand

\frac{x ^{6} - x}{x ^{3}} : x^{3} - \frac{1}{x ^{2}}

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

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

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

= \frac{x ^{4}}{4} - (- \frac{1}{x})

Simplify

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

Add a constant to the solution

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

\int \frac{x + 2}{x ^{2} + 7 x - 8} d x

2.87 y^{^{''}} + 10.5 y^{^{'}} - 1.83 y = 0

y^{^{'}} = y^{2} (1 + e^{x})

\frac{d}{d x} (\frac{4 x ^{5} + 7}{x ^{3}})

y^{^{''}} - 2 y^{^{'}} + y = \frac{( - 6 e ^{t} )}{t ^{2} + 1}

What is the integral of (x^6-x)/(x^3) ?
The integral of (x^6-x)/(x^3) is (x^4)/4+1/x+C