What is the integral of (sqrt(x^4+x^{-4)+2})/(x^3) ?

The integral of (sqrt(x^4+x^{-4)+2})/(x^3) is ln|x|-1/(4x^4)+C

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integral of (sqrt(x^4+x^{-4)+2})/(x^3)

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

ln ∣ x ∣ - \frac{1}{4 x ^{4}} + C

Solution steps

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

Simplify

x^{4} : x^{3} x

= \int \frac{x ^{3} x + x ^{- 4} + 2}{x ^{3}} d x

Simplify

\frac{x ^{3} x + x ^{- 4} + 2}{x ^{3}} : \frac{( x ^{2} + 2 x + 1 ) ( x ^{2} - 2 x + 1 )}{x ^{5}}

= \int \frac{( x ^{2} + 2 x + 1 ) ( x ^{2} - 2 x + 1 )}{x ^{5}} d x

Expand

\frac{( x ^{2} + 2 x + 1 ) ( x ^{2} - 2 x + 1 )}{x ^{5}} : \frac{1}{x} + \frac{1}{x ^{5}}

= \int \frac{1}{x} + \frac{1}{x ^{5}} 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} d x + \int \frac{1}{x ^{5}} d x

\int \frac{1}{x} d x = ln ∣ x ∣

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

= ln ∣ x ∣ - \frac{1}{4 x ^{4}}

Add a constant to the solution

= ln ∣ x ∣ - \frac{1}{4 x ^{4}} + C

4 x + 84

arcsec (6 x)

\frac{\partial}{\partial x} (e^{x} g)

36 y^{^{''}} - 36 y^{^{'}} - 7 y = 0

x \to a lim (f (x))

What is the integral of (sqrt(x^4+x^{-4)+2})/(x^3) ?
The integral of (sqrt(x^4+x^{-4)+2})/(x^3) is ln|x|-1/(4x^4)+C