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

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

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integral of (1/(x^2)+2/(x^3)+3/(x^4))

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

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

Solution steps

\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}}

Add a constant to the solution

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

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