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

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

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

integral

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

x + \frac{1}{x} - \frac{1}{2 x ^{2}} + C

Solution steps

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

Expand

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

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

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

\int 1 d x = x

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

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

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

Simplify

= x + \frac{1}{x} - \frac{1}{2 x ^{2}}

Add a constant to the solution

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

in t e g r a l 32 e^{- 4} x

in t e g r a l \frac{( x + 3 )}{( 2 x + 1 )}

in t e g r a l 2 x sin (x^{2} + 1)

in t e g r a l \frac{( x + 3 )}{( x ^{2} - x + 6 )}

in t e g r a l \frac{t ^{2}}{2} + \frac{t ^{3}}{3} + \frac{t ^{4}}{4}

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