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

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

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

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

x + \frac{2}{x} - \frac{7}{3 x ^{3}} + C

Solution steps

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

Expand

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

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

\int 1 d x = x

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

\int \frac{7}{x ^{4}} d x = - \frac{7}{3 x ^{3}}

= x - (- \frac{2}{x}) - \frac{7}{3 x ^{3}}

Simplify

= x + \frac{2}{x} - \frac{7}{3 x ^{3}}

Add a constant to the solution

= x + \frac{2}{x} - \frac{7}{3 x ^{3}} + C

\int_{0}^{1} \frac{x ^{4} + 8}{x} d x

\int x e^{x^{2} + 1} d x

(1 + 8 x^{2}) (x - x^{2})

d x + e^{4 x} d y = 0

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