на фото..............

Раскладываем знаменатель на множители:

int (dx* (2x+5)/ ( x^2*(x-1) +2*(x-1) ) ) = int (  dx*(2x+5)/(x^2+2)*(x-1) )=

int( dx* ((2x-2) +7)/(x^2+2)*(x-1) ) = 2*int(dx/(x^2+2))  +7*int(dx/(x^2+2)*(x-1) )

Решаю без метода неопределенных коэффициентов просто подгоняю  числитель под нужный:

1/(x-1)   +1/(x^2+2) =(x^2+2 +x-1)/(x^2+2)*(x-1) =  (x^2+x+1)/(x^2+2)*(x-1)=

=  ((x^2-x) +(2x-2) +3)/(x^2+2)*(x-1)  = x/(x^2+2)  +  2/(x^2+2)  +3/(x^2+2)*(x-1)

1/(x^2+2)*(x-1)=  1/3  *(1/(x-1) +1/(x^2+2) -x/(x^2+2) -2/(x^2+2) )=

=1/3 *( 1/(x-1) -1/(x^2+2)  -x/(x^2+2) )

2*int(dx/(x^2+2))  +7*int(dx/(x^2+2)*(x-1) )=  

=7/3*int(dx/(x-1) ) -1/3 *int(dx/(x^2+2))  -7/3*int(dx*x/(x^2+2))=

7/3 *ln(x-1) +c  -√2/6 *int ((1/√2)*dx/(1+(x/√2)^2 )  -7/6 *int (2x*dx/(x^2+2))=

=7/3*ln(x-1) -√2/6 *arctg(x/√2) -7/6*ln(x^2+2) +c=

7/6 *( 2*ln(x-1) -ln(x^2+2) ) -√2/6*arctg(x/√2) +c=7/6*ln((x-1)^2/(x^2+2)) +c-√2/6*arctg(x/√2)= 7/6 *ln ((x-1)^2/(x^2+2) ) -√2/6 *arctg(x*√2/2) +c

Ответ:( 7 *ln ( (x-1)^2 /(x^2+2) )-√2arctg(x*√2/2 ) )/6  +c