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}}

Add a constant to the solution

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

\frac{d x}{d t} = 8 + 4 sin (\frac{t}{4}) - \frac{x}{100}, x (0) = 70

f (x) = \frac{x}{2 x - 8}, at x = 3

x \to \infty lim (\frac{5}{x} - 4)

y = \frac{ln ( x )}{x ^{9}}

\int \frac{1}{2} ln (x) d 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