partial-fraction-with-detail-step-by-step-from-4x-3-x-2-1-x-2-2-3-

Question Number 125867 by joki last updated on 14/Dec/20

partial fraction with detail step by step from

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

Answered by mathmax by abdo last updated on 14/Dec/20

F (x) = \frac{{4 x}^{3} + x^{2} + 1}{{(x^{2} - 2)}^{3}} \Rightarrow F (x) = \frac{{4 x}^{3} + x^{2} + 1}{{(x - \sqrt{2})}^{3} {(x + \sqrt{2})}^{3}}

= \frac{a}{x - \sqrt{2}} + \frac{b}{{(x - \sqrt{2})}^{2}} + \frac{c}{{(x - \sqrt{2})}^{3}} + \frac{d}{x + \sqrt{2}} + \frac{e}{{(x + \sqrt{2})}^{2}} + \frac{d}{{(x + \sqrt{2})}^{3}}

we can know c and d immediatly

c = {(x - 2)}^{3} F (x) ∣_{x = \sqrt{2}} = \frac{4 {(\sqrt{2})}^{3} + 2 + 1}{{(2 \sqrt{2})}^{3}} = \frac{8 \sqrt{2} + 3}{16 \sqrt{2}}

d = (x + \sqrt{2}) F (x) ∣_{x = - \sqrt{2}} = \frac{4 {(- \sqrt{2})}^{3} + 2 + 1}{{(- 2 \sqrt{2})}^{3}} = \dots .

\lim_{x \to + \infty} xF (x) = 0 = a + d \Rightarrow d = - a

we can calculate F (0), F (1) and F (- 1) to get a system after

we slve it \dots . any way there is a lots of calculus \dots

Commented by mathmax by abdo last updated on 14/Dec/20

another way by integral we have

F (x) = \frac{d}{dx} (\int F (x) dx) we have \int F (x) dx = \int \frac{{4 x}^{3} + x^{2} + 1}{{(x^{2} - 2)}^{3}} dx

= \int \frac{{4 x}^{3} + x^{2} + 1}{{(x - \sqrt{2})}^{3} {(x + \sqrt{2})}^{3}} dx = \int \frac{{4 x}^{3} + x^{2} + 1}{{(\frac{x - \sqrt{2}}{x + \sqrt{2}})}^{3} {(x + \sqrt{2})}^{6}} dx

we do the changement \frac{x - \sqrt{2}}{x + \sqrt{2}} = t \Rightarrow x - \sqrt{2} = tx + \sqrt{2} t \Rightarrow

(1 - t) x = \sqrt{2} t + \sqrt{2} \Rightarrow x = \frac{\sqrt{2} t + \sqrt{2}}{1 - t} \Rightarrow \frac{dx}{dt} = \frac{\sqrt{2} (1 - t) + \sqrt{2} t + \sqrt{2}}{{(1 - t)}^{2}}

= \frac{2 \sqrt{2}}{{(t - 1)}^{2}} and x + \sqrt{2} = \frac{\sqrt{2} t + \sqrt{2}}{1 - t} + \sqrt{2} = \frac{2 \sqrt{2}}{1 - t} \Rightarrow

\int F (x) dx = \int \frac{4 {(\frac{\sqrt{2} t + \sqrt{2}}{1 - t})}^{3} + {(\frac{\sqrt{2} t + \sqrt{2}}{1 - t})}^{2} + 1}{t^{3} {(\frac{2 \sqrt{2}}{1 - t})}^{6}} \times \frac{2 \sqrt{2}}{{(t - 1)}^{2}} dt

= \frac{1}{{(2 \sqrt{2})}^{5}} \int \frac{1}{{(1 - t)}^{3}} (4 {(\sqrt{2} t + \sqrt{2})}^{3} + (1 - t) {(\sqrt{2} t + \sqrt{2})}^{2} + {(1 - t)}^{3}) 2 \sqrt{2} {(1 - t)}^{4} \frac{dt}{t^{3}}

= - \frac{1}{{(2 \sqrt{2})}^{5}} \int (t - 1) {4 {(\sqrt{2} t + \sqrt{2})}^{3} + (1 - t) {(\sqrt{2} t + \sqrt{2})}^{2} + {(1 - t)}^{3}) dx

its eazy to find this integral so you get decompositino of

\int F (x) dx after we derivate the integral this method is sure \dots

Commented by joki last updated on 15/Dec/20

thank you sir for your solution .

Commented by mathmax by abdo last updated on 16/Dec/20

you are welcome