What is the integral of (x^4-3x^2+5)/(x^4) ?

The integral of (x^4-3x^2+5)/(x^4) is x+3/x-5/(3x^3)+C

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integral of (x^4-3x^2+5)/(x^4)

\int \frac{x ^{4} - 3 x ^{2} + 5}{x ^{4}} d x

x + \frac{3}{x} - \frac{5}{3 x ^{3}} + C

Solution steps

\int \frac{x ^{4} - 3 x ^{2} + 5}{x ^{4}} d x

Expand

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

= \int 1 - \frac{3}{x ^{2}} + \frac{5}{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 1 d x - \int \frac{3}{x ^{2}} d x + \int \frac{5}{x ^{4}} d x

\int 1 d x = x

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

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

= x - (- \frac{3}{x}) - \frac{5}{3 x ^{3}}

Simplify

= x + \frac{3}{x} - \frac{5}{3 x ^{3}}

Add a constant to the solution

= x + \frac{3}{x} - \frac{5}{3 x ^{3}} + C

x \to - 2 lim (\frac{x + 2}{2 x})

\int x sin^{2} (x^{2}) d x

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

x \to 0 lim (\frac{60}{x ^{4}})

\frac{d}{d x} (\frac{sec ( x )}{x})

What is the integral of (x^4-3x^2+5)/(x^4) ?
The integral of (x^4-3x^2+5)/(x^4) is x+3/x-5/(3x^3)+C