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

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

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

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

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

Solution steps

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

\int \frac{x}{x ^{2} - 4} d x = \frac{1}{2} ln x^{2} - 4

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

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

Add a constant to the solution

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

\int 6 sin (2 x) - 3 cos (3 x) d x

x \to 1 lim ((x + 2)^{4} (4 x^{2}))

\frac{d}{d x} (\frac{3 csc ( x )}{4 ^{x}})

\int \frac{5}{2} x d x

\frac{\partial}{\partial x} (sec (17 x))

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