What is the integral of (1+x^2)/(x^4) ?

The integral of (1+x^2)/(x^4) is -1/(3x^3)-1/x+C

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integral of (1+x^2)/(x^4)

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

- \frac{1}{3 x ^{3}} - \frac{1}{x} + C

Solution steps

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

Expand

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

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

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

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

= - \frac{1}{3 x ^{3}} - \frac{1}{x}

Add a constant to the solution

= - \frac{1}{3 x ^{3}} - \frac{1}{x} + C

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What is the integral of (1+x^2)/(x^4) ?
The integral of (1+x^2)/(x^4) is -1/(3x^3)-1/x+C