What is the integral of (x^2+1)/(x^6) ?

The integral of (x^2+1)/(x^6) is -1/(3x^3)-1/(5x^5)+C

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integral of (x^2+1)/(x^6)

\int \frac{x ^{2} + 1}{x ^{6}} d x

- \frac{1}{3 x ^{3}} - \frac{1}{5 x ^{5}} + C

Solution steps

\int \frac{x ^{2} + 1}{x ^{6}} d x

Expand

\frac{x ^{2} + 1}{x ^{6}} : \frac{1}{x ^{4}} + \frac{1}{x ^{6}}

= \int \frac{1}{x ^{4}} + \frac{1}{x ^{6}} 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 ^{6}} d x

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

\int \frac{1}{x ^{6}} d x = - \frac{1}{5 x ^{5}}

= - \frac{1}{3 x ^{3}} - \frac{1}{5 x ^{5}}

Add a constant to the solution

= - \frac{1}{3 x ^{3}} - \frac{1}{5 x ^{5}} + C

\int \frac{1}{- 3 x ^{2}} d x

\int (5 y) d y

tan^{4} (x)

\int \frac{2}{3 - 4 x} d x

\int \frac{( e ^{x} )}{( 2 - e ^{x} ) ^{2}} d x

What is the integral of (x^2+1)/(x^6) ?
The integral of (x^2+1)/(x^6) is -1/(3x^3)-1/(5x^5)+C