What is the integral of (x^4-7x^2+5)/(x^4) ?

The integral of (x^4-7x^2+5)/(x^4) is x+7/x-5/(3x^3)+C

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integral of (x^4-7x^2+5)/(x^4)

\int \frac{x ^{4} - 7 x ^{2} + 5}{x ^{4}} d x

x + \frac{7}{x} - \frac{5}{3 x ^{3}} + C

Solution steps

\int \frac{x ^{4} - 7 x ^{2} + 5}{x ^{4}} d x

Expand

\frac{x ^{4} - 7 x ^{2} + 5}{x ^{4}} : 1 - \frac{7}{x ^{2}} + \frac{5}{x ^{4}}

= \int 1 - \frac{7}{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{7}{x ^{2}} d x + \int \frac{5}{x ^{4}} d x

\int 1 d x = x

\int \frac{7}{x ^{2}} d x = - \frac{7}{x}

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

= x - (- \frac{7}{x}) - \frac{5}{3 x ^{3}}

Simplify

= x + \frac{7}{x} - \frac{5}{3 x ^{3}}

Add a constant to the solution

= x + \frac{7}{x} - \frac{5}{3 x ^{3}} + C

\int_{0}^{3} \frac{1}{( x - 1 ) ^{\frac{2}{3}}} d x

y^{^{'}} = \frac{t ^{3} + 19 t ^{3} y}{t}

\frac{\partial}{\partial y} (- yz)

x \to - 7 lim (\frac{1}{x})

x \to 9 lim (cos (\frac{π x}{3}))

What is the integral of (x^4-7x^2+5)/(x^4) ?
The integral of (x^4-7x^2+5)/(x^4) is x+7/x-5/(3x^3)+C