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

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

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

\int \frac{( x ^{2} + 1 ) ( x ^{4} - x ^{2} + 1 )}{2 x ^{3}} d x

\frac{1}{2} (\frac{x ^{4}}{4} - \frac{1}{2 x ^{2}}) + C

Solution steps

\int \frac{( x ^{2} + 1 ) ( x ^{4} - x ^{2} + 1 )}{2 x ^{3}} d x

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= \frac{1}{2} \cdot \int \frac{( x ^{2} + 1 ) ( x ^{4} - x ^{2} + 1 )}{x ^{3}} d x

Expand

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

= \frac{1}{2} \cdot \int x^{3} + \frac{1}{x ^{3}} 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

= \frac{1}{2} (\int x^{3} d x + \int \frac{1}{x ^{3}} d x)

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

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

= \frac{1}{2} (\frac{x ^{4}}{4} - \frac{1}{2 x ^{2}})

Add a constant to the solution

= \frac{1}{2} (\frac{x ^{4}}{4} - \frac{1}{2 x ^{2}}) + C

(1 + 9 x^{2}) (x + x^{2})

\int \frac{2}{4 - 9 x ^{2}} d x

x \to - 4 lim (6 - 3 x)

y = x^{2} 2^{x^{2}}

(x + 3 t) \frac{d x}{d t} + 3 x - t = 0

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