①∫dx/[x(ax+b)] =a*∫[1/(ax)*(ax+b)]dx =(1/b)∫[(1/ax)-(1/ax+b)]d(ax) =(1/b)*[ln|ax|-ln|ax+b|]+C =(1/b)*ln|ax/(ax+b)|+C
x²/(ax+b) =[(x+b/a)(x-b/a)+b²/a²]/[a(x+b/a)] =(x-b/a)/a+b²/(a³x+a²b) 所以原式=∫(x-b/a)/adx+∫b²/(a³x+a²b)/adx =(x²/2-bx/a)/a+b²/a^4*ln|a³x+a²b|+C
是0.5吧?最后那个? 等式为=1+1/4*1/(3/2)+0.1=1+1/6+0.1=1.1+1/6
先算除法得5/8,2/9+5/8+3/8=2/9+1=1 2/9(一又九分之二)