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

The integral of ((x^3-4x))/((x^2+1)^2) is 1/2 ln|x^2+1|+5/(2(x^2+1))+C

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

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

\frac{1}{2} ln x^{2} + 1 + \frac{5}{2 ( x ^{2} + 1 )} + C

Solution steps

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

Take the partial fraction of

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

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

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

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

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

Simplify

= \frac{1}{2} ln x^{2} + 1 + \frac{5}{2 ( x ^{2} + 1 )}

Add a constant to the solution

= \frac{1}{2} ln x^{2} + 1 + \frac{5}{2 ( x ^{2} + 1 )} + C

\int 7 x^{4} - 5 x^{3} + 4 x^{2} + C d x

(x + 2) y^{^{'}} - y = \frac{3}{x + 2}

x \to \infty + lim (\frac{1}{x})

(\frac{t}{y}) d y + (1 + ln (y)) d t = 0

\int \frac{1}{6} sec (x) tan (x) d x

What is the integral of ((x^3-4x))/((x^2+1)^2) ?
The integral of ((x^3-4x))/((x^2+1)^2) is 1/2 ln|x^2+1|+5/(2(x^2+1))+C