calculate-1-dx-4x-2-1-3-

Question Number 115927 by mathmax by abdo last updated on 29/Sep/20

calculate \int_{1}^{+ \infty} \frac{dx}{{({4 x}^{2} - 1)}^{3}}

Answered by 1549442205PVT last updated on 29/Sep/20

\frac{2}{{4 x}^{2} - 1} = \frac{1}{2 x - 1} - \frac{1}{2 x + 1} . Put \frac{1}{2 x - 1} = a,

\frac{1}{2 x + 1} = b \Rightarrow 2 ab = a - b

\frac{1}{{({4 x}^{2} - 1)}^{3}} = \frac{1}{8} {(\frac{2}{{4 x}^{2} - 1})}^{3} = \frac{1}{8} {(a - b)}^{3}

= \frac{1}{8} [a^{3} - b^{3} - 3 ab (a - b)] = \frac{1}{8} [a^{3} - b^{3} - \frac{3}{2} {(a - b)}^{2}]

\frac{1}{8} [a^{3} - b^{3} - \frac{3}{2} a^{2} - \frac{3}{2} b^{2} + \frac{3}{2} (a - b)]

= \frac{1}{8} a^{3} - \frac{1}{8} b^{3} - \frac{3}{16} a^{2} - \frac{3}{16} b^{2} + \frac{3}{16} a - \frac{3}{16} b

I = \int_{1}^{+ \infty} \frac{dx}{{({4 x}^{2} - 1)}^{3}}

= \int_{1}^{\infty} (\frac{1}{8 {(2 x - 1)}^{3}} - \frac{1}{8 {(2 x + 1)}^{3}} - \frac{3}{16 {(2 x - 1)}^{2}} - \frac{3}{16 {(2 x + 1)}^{2}} + \frac{3}{16 (2 x - 1)} - \frac{3}{16 (2 x + 1)}) dx

= \frac{1}{16} \int_{1}^{\infty} \frac{d (2 x - 1)}{{(2 x - 1)}^{3}} - \frac{1}{16} \int_{1}^{\infty} \frac{d (2 x + 1)}{{(2 x + 1)}^{3}}

- \frac{3}{32} \int_{1}^{\infty} \frac{d (2 x - 1)}{{(2 x - 1)}^{2}} - \frac{3}{32} \int_{1}^{\infty} \frac{d (2 x + 1)}{{(2 x + 1)}^{2}}

+ \frac{3}{32} \int \frac{d (2 x - 1)}{(2 x - 1)} - \frac{3}{32} \int_{1}^{\infty} \frac{d (2 x + 1)}{(2 x + 1)}

= {\frac{- 1}{32 {(2 x - 1)}^{2}} + \frac{1}{32 {(2 x + 1)}^{2}} + \frac{3}{32 (2 x - 1)} + \frac{3}{32 (2 x + 1)} + \frac{3}{32} \ln ∣ \frac{2 x - 1}{2 x + 1} ∣}_{1}^{\infty}

= \frac{1}{32} - \frac{1}{288} - \frac{3}{32} - \frac{1}{32} - \frac{3}{32} \ln (\frac{1}{3}) = \frac{- 7}{72} + \frac{3}{32} \ln 3

Answered by mathmax by abdo last updated on 29/Sep/20

I = \int_{1}^{+ \infty} \frac{dx}{{({4 x}^{2} - 1)}^{3}} \Rightarrow I = \int_{1}^{\infty} \frac{dx}{{(2 x - 1)}^{3} {(2 x + 1)}^{3}}

= \int_{1}^{\infty} \frac{dx}{{(\frac{2 x - 1}{2 x + 1})}^{3} {(2 x + 1)}^{6}} we do the changement \frac{2 x - 1}{2 x + 1} = t \Rightarrow

2 x - 1 = 2 tx + t \Rightarrow (2 - 2 t) x = t + 1 \Rightarrow x = \frac{t + 1}{2 - 2 t} \Rightarrow

\frac{dx}{dt} = \frac{2 - 2 t - (- 2) (t + 1)}{{(2 - 2 t)}^{2}} = \frac{1}{4 {(1 - t)}^{2}} and 2 x + 1 = \frac{2 t + 2}{2 - 2 t} + 1

= \frac{2 t + 2 + 2 - 2 t}{2 - 2 t} = \frac{4}{2 - 2 t} = \frac{2}{1 - t} \Rightarrow

I = \int_{\frac{1}{3}}^{1} \frac{1}{t^{3} {(\frac{2}{1 - t})}^{6}} \times \frac{dt}{4 {(1 - t)}^{2}} = \frac{1}{4} \int_{\frac{1}{3}}^{1} \frac{{(t - 1)}^{6}}{2^{6} . t^{3} . {(t - 1)}^{2}} dt

= \frac{1}{2^{8}} \int_{\frac{1}{3}}^{1} \frac{{(t - 1)}^{4}}{t^{3}} dt \Rightarrow 2^{8} . I = \int_{\frac{1}{3}}^{1} \frac{\sum_{k = 0}^{4} C_{4}^{k} t^{k} {(- 1)}^{4 - k}}{t^{3}} dt

= \sum_{k = 0}^{4} {(- 1)}^{k} C_{4}^{k} \int_{\frac{1}{3}}^{1} t^{k - 3} dt

= \sum_{k = 0 and k \neq 2}^{4} {(- 1)}^{k} C_{4}^{k} {[\frac{1}{k - 2} t^{k - 2}]}_{\frac{1}{3}}^{1} + C_{4}^{2} {[lnt]}_{\frac{1}{3}}^{1} \Rightarrow

I = \frac{1}{2^{8}} (\sum_{k = 0 and k \neq 2}^{4} \frac{{(- 1)}^{k} C_{4}^{k}}{k - 2} (1 - \frac{1}{3^{k - 2}}) + C_{4}^{2} \ln (3))