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

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

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

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

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

Solution steps

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

\int \frac{1}{x} d x = ln ∣ x ∣

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

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

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

= ln ∣ x ∣ - \frac{1}{x} - \frac{1}{2 x ^{2}} - \frac{1}{3 x ^{3}}

Add a constant to the solution

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

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

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

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

\int x sin (x + 2) d x

\int (sin (x) + e^{x}) d x

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