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Answer (1)
Respuesta: ∫ ((x)/((4x-1)(x+2))dx = (-1/28)ln|(4x-1)|-(2/7)ln|(x+2)|+C ∫ ((x)/((4x-1)(x+2))dx ; ((x)/((4x-1)(x+2)) = (A+B))/((4x-1)(x+2)) === > ((x)/((4x+1)(X-23)) = ((A)/(4x-1))+((B/(x+2)) === > x = A(x+2)+B(4x-1) ==== > x = x(A+4B)+2A-B BEntonces :x = x(A+4B) ==== > A+4B = 10 = 2A-B ===== > 2A-B = 0 Método de Igualación :1 ) Se despeja la incógnita '' B '' , en la ecuación '' 2A-B = 0 "' : 2A-B = 02A = BB = 2A2 ) Se despeja la incógnita '' B '' , en la ecuación '' A+4B = 1 '' :A+4B = 1A-1 = 4B(A-1)/4 = BB = (A-1)/43 ) Se igualan las ecuaciones resultantes '' B = 2A " y '' B = (A-1)/4 '' : 2A = (A-1)/44(2A) = A-18A = A-17A = - 1A = - 1/74 ) Se sustituye el valor de la incógnita '' A '' , que es - 1/7 , en la ecuación '' B = 2A '' :B = 2(-1/7)B = - 2/7Por lo tanto :∫((x)/((4x-1)(x+2))dx = ∫((-1/7)/(4x+1))dx+∫ ((-2/7)/(x+2))dx = (-1/7)∫((1)/(4x+1))dx-(2/7)∫((1)/(x+2))dxEntonces :∫ ((x)/((4x-1)(x+2))dx = (-1/7)∫((1)/((4x-1))dx-(2/7)∫((1)/(x+2))dx(-1/7)∫((1)/((4x-1))dx = (-1/28)ln|(4x-1)|+C(-1/7)∫((1)/((4x-1))dx;= (-1/7)∫((1)/((4x-1))dx; (4x-1) = j === > (4x-1)' = j' === > 4 = dj/dx === > dx = (1/4)dj= ((-1/7)×(1/4))∫((1)/(4x-1))dj= (-1/28)∫((1)/(4x-1))dj ; (4x-1) = j= (-1/28)∫((1)/(j))dj ; ∫((1)/(j))dj = ln|(j)|= (-1/28)ln|(j)| ; j = (4x-1)= (-1/28)ln|(4x-1)|= (-1/28)ln|(4x-1)|+C-(2/7)∫((1)/(x+2))dx = -(2/7)ln|(x+2)|+C-(2/7)∫((1)/(x+2))dx ;= -(2/7)∫((1)/(x+2))dx ; (x+2) = h ==== > ((x+2))' = h' ==== > 1 = dh/dx ==== > dx = dh= -(2/7)∫((1)/(x+2))dh ; (x+2) = h= -(2/7)∫((1)/(h))dh ; ∫((1)/(h))dh = ln|(h)|= -(2/7)ln|(h)| ; h = (x+2)= -(2/7)ln|(x+2)|= -(2/7)ln|(x+2)|+CPor tanto : ∫((x)/((4x-1)(x+2))dx = (-1/28)ln|(4x-1)|-(2/7)ln|(x+2)|+CR// La respuesta de calcular ∫((x)/((4x-1)(x+2)))dx es (-1/28)ln|(4x-1)|-(2/7)ln|(x+2)|+C .∫((1)/((x+a)(x+b))dx = ((1)/(b-a))(ln|((x+a)/(x+b))|)+C∫((1)/((x+a)(x+b))dx ; ((1)/((x+a)(x+b)) = (A+B)/((x+a)(x+b)) ===== > ((1)/((x+a)(x+b))) = ((A)/(x+a))+((B)/(x+b))==== > ∫((1)/((x+a)(x+b))dx = ∫((A)/((x+a))dx+∫((B)/(x+b))dx1 = A(x+b)+B(x+a)1 = Ab+Ba+x(A+B)0x = x(A+B) ==== > A+B = 0Ab+Ba = 1 ==== > Ab+Ba = 1 Método de Sustitución :1 ) Se despeja la variable '' B '' , en la ecuación '' A+B = 0 '' :A+B = 0B = - A2 ) Se sustituye el resultado de despejar la variable '' B '' , el cual es - A , en la ecuación '' Ab+Ba = 1 '' :Ab+Ba = 1 ; B = - AAb+(-A)a = 1Ab-Aa = 1A(b-a) = 1A = ((1)/(b-a))Posterior a eso :B = - A y como A = ((1)/(b-a)) === > B = -((1)/(b-a))Por lo cual :∫((1)/((x+a)(x+b))dx = ((1)/(b-a))(∫((1)/(x+a))dx-∫((1)/(x+b)))∫((1)/((x+a)(x+b))dx = ((1)/(b-a))(ln|(x+a)|-ln|(x+b)|)∫((1)/((x+a)(x+b))dx = ((1)/(b-a))(ln|((x+a)/(x+b))|)+CR// Por ende , el resultado de calcular ∫((1)/((x+a)(x+b))dx es ((1)/(b-a))(ln|((x+a)/(x+b))|)+C
