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

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

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

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

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

Solution steps

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

Expand

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

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

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

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

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

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

Simplify

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

Add a constant to the solution

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

x = e^{y}, y = 0, x = 1, x = e^{2}

\int e^{- 2 x^{2}} x d x

\int \frac{x ^{3}}{3 x ^{2} + 7} d x

\int 12 e^{x} d x

\frac{6 t d y}{d t} + y = t^{6}

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