calculate ∫_1 ^(+∞) (dx/((4x^2 −1)^3 ))


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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
	$(2/(4x^2 −1))=(1/(2x−1))−(1/(2x+1)).Put (1/(2x−1))=a, (1/(2x+1))=b⇒2ab=a−b (1/((4x^2 −1)^3 ))=(1/8)((2/(4x^2 −1)))^3 =(1/8)(a−b)^3 =(1/8)[a^3 −b^3 −3ab(a−b)]=(1/8)[a^3 −b^3 −(3/2)(a−b)^2 ] (1/8)[a^3 −b^3 −(3/2)a^2 −(3/2)b^2 +(3/2)(a−b)] =(1/8)a^3 −(1/8)b^3 −(3/(16))a^2 −(3/(16))b^2 +(3/(16))a−(3/(16))b I=∫_1 ^(+∞) (dx/((4x^2 −1)^3 )) =∫_1 ^∞ ((1/(8(2x−1)^3 ))−(1/(8(2x+1)^3 ))−(3/(16(2x−1)^2 ))−(3/(16(2x+1)^2 ))+(3/(16(2x−1)))−(3/(16(2x+1))))dx =(1/(16))∫_1 ^∞ ((d(2x−1))/((2x−1)^3 ))−(1/(16))∫_1 ^∞ ((d(2x+1))/((2x+1)^3 )) −(3/(32))∫_1 ^∞ ((d(2x−1))/((2x−1)^2 ))−(3/(32))∫_1 ^∞ ((d(2x+1))/((2x+1)^2 )) +(3/(32))∫((d(2x−1))/((2x−1)))−(3/(32))∫_1 ^∞ ((d(2x+1))/((2x+1))) ={((−1)/(32(2x−1)^2 ))+(1/(32(2x+1)^2 ))+(3/(32(2x−1)))+(3/(32(2x+1)))+(3/(32))ln∣((2x−1)/(2x+1))∣}_1 ^∞ =(1/(32))−(1/(288))−(3/(32))−(1/(32))−(3/(32))ln((1/3))=((−7)/(72))+(3/(32))ln3$
	$\frac{2}{{4 x}^{2} - 1} = \frac{1}{2 x - 1} - \frac{1}{2 x + 1} . Put \frac{1}{2 x - 1} = a,$ $\frac{1}{2 x + 1} = b \Rightarrow 2 ab = a - b$ $\frac{1}{{({4 x}^{2} - 1)}^{3}} = \frac{1}{8} {(\frac{2}{{4 x}^{2} - 1})}^{3} = \frac{1}{8} {(a - b)}^{3}$ $= \frac{1}{8} [a^{3} - b^{3} - 3 ab (a - b)] = \frac{1}{8} [a^{3} - b^{3} - \frac{3}{2} {(a - b)}^{2}]$ $\frac{1}{8} [a^{3} - b^{3} - \frac{3}{2} a^{2} - \frac{3}{2} b^{2} + \frac{3}{2} (a - b)]$ $= \frac{1}{8} a^{3} - \frac{1}{8} b^{3} - \frac{3}{16} a^{2} - \frac{3}{16} b^{2} + \frac{3}{16} a - \frac{3}{16} b$ $I = \int_{1}^{+ \infty} \frac{dx}{{({4 x}^{2} - 1)}^{3}}$ $= \int_{1}^{\infty} (\frac{1}{8 {(2 x - 1)}^{3}} - \frac{1}{8 {(2 x + 1)}^{3}} - \frac{3}{16 {(2 x - 1)}^{2}} - \frac{3}{16 {(2 x + 1)}^{2}} + \frac{3}{16 (2 x - 1)} - \frac{3}{16 (2 x + 1)}) dx$ $= \frac{1}{16} \int_{1}^{\infty} \frac{d (2 x - 1)}{{(2 x - 1)}^{3}} - \frac{1}{16} \int_{1}^{\infty} \frac{d (2 x + 1)}{{(2 x + 1)}^{3}}$ $- \frac{3}{32} \int_{1}^{\infty} \frac{d (2 x - 1)}{{(2 x - 1)}^{2}} - \frac{3}{32} \int_{1}^{\infty} \frac{d (2 x + 1)}{{(2 x + 1)}^{2}}$ $+ \frac{3}{32} \int \frac{d (2 x - 1)}{(2 x - 1)} - \frac{3}{32} \int_{1}^{\infty} \frac{d (2 x + 1)}{(2 x + 1)}$ $= {\frac{- 1}{32 {(2 x - 1)}^{2}} + \frac{1}{32 {(2 x + 1)}^{2}} + \frac{3}{32 (2 x - 1)} + \frac{3}{32 (2 x + 1)} + \frac{3}{32} \ln ∣ \frac{2 x - 1}{2 x + 1} ∣}_{1}^{\infty}$ $= \frac{1}{32} - \frac{1}{288} - \frac{3}{32} - \frac{1}{32} - \frac{3}{32} \ln (\frac{1}{3}) = \frac{- 7}{72} + \frac{3}{32} \ln 3$
	Answered by mathmax by abdo last updated on 29/Sep/20

	$I = \int_{1}^{+ \infty} \frac{dx}{{({4 x}^{2} - 1)}^{3}} \Rightarrow I = \int_{1}^{\infty} \frac{dx}{{(2 x - 1)}^{3} {(2 x + 1)}^{3}}$ $= \int_{1}^{\infty} \frac{dx}{{(\frac{2 x - 1}{2 x + 1})}^{3} {(2 x + 1)}^{6}} we do the changement \frac{2 x - 1}{2 x + 1} = t \Rightarrow$ $2 x - 1 = 2 tx + t \Rightarrow (2 - 2 t) x = t + 1 \Rightarrow x = \frac{t + 1}{2 - 2 t} \Rightarrow$ $\frac{dx}{dt} = \frac{2 - 2 t - (- 2) (t + 1)}{{(2 - 2 t)}^{2}} = \frac{1}{4 {(1 - t)}^{2}} and 2 x + 1 = \frac{2 t + 2}{2 - 2 t} + 1$ $= \frac{2 t + 2 + 2 - 2 t}{2 - 2 t} = \frac{4}{2 - 2 t} = \frac{2}{1 - t} \Rightarrow$ $I = \int_{\frac{1}{3}}^{1} \frac{1}{t^{3} {(\frac{2}{1 - t})}^{6}} \times \frac{dt}{4 {(1 - t)}^{2}} = \frac{1}{4} \int_{\frac{1}{3}}^{1} \frac{{(t - 1)}^{6}}{2^{6} . t^{3} . {(t - 1)}^{2}} dt$ $= \frac{1}{2^{8}} \int_{\frac{1}{3}}^{1} \frac{{(t - 1)}^{4}}{t^{3}} dt \Rightarrow 2^{8} . I = \int_{\frac{1}{3}}^{1} \frac{\sum_{k = 0}^{4} C_{4}^{k} t^{k} {(- 1)}^{4 - k}}{t^{3}} dt$ $= \sum_{k = 0}^{4} {(- 1)}^{k} C_{4}^{k} \int_{\frac{1}{3}}^{1} t^{k - 3} dt$ $= \sum_{k = 0 and k \neq 2}^{4} {(- 1)}^{k} C_{4}^{k} {[\frac{1}{k - 2} t^{k - 2}]}_{\frac{1}{3}}^{1} + C_{4}^{2} {[lnt]}_{\frac{1}{3}}^{1} \Rightarrow$ $I = \frac{1}{2^{8}} (\sum_{k = 0 and k \neq 2}^{4} \frac{{(- 1)}^{k} C_{4}^{k}}{k - 2} (1 - \frac{1}{3^{k - 2}}) + C_{4}^{2} \ln (3))$
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