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^{2} \frac{d y}{d x} - x = 1 + y + x y

f (x) = 12 x

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

\frac{1}{1 - x}, 9

\frac{d}{d x} (16 sin (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