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

The integral of ((x-2)/(x(x^2+2))) is 1/(sqrt(2))arctan(x/(sqrt(2)))+1/2 ln|x^2+2|-ln(x)+C

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integral of ((x-2)/(x(x^2+2)))

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

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

Solution steps

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

Expand

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

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

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

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

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

Simplify

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

Add a constant to the solution

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

\int \frac{1}{( u + 1 ) ^{2}} d u

x^{2} y^{^{'}} = 1 - x^{2} + y^{2} - x^{2} y^{2}

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

\frac{d y}{d x} = - sin^{2} (x + y)

\int_{- 1}^{1} x + 2 d x

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