Question

求解两道高数题，急

1、设y=x^5/x^2+7x+12,求y的n阶导
2、设f(x)连续，且满足f(xy)=f(x)+f(y)，证明：
∫(0到1)f（1+x）/(1+x^2) dx=π/8*f(π)

Answer 1

y=x^5/(x^2+7x+12)=x^5/(x+3)(x+5)
=(x^5+3x^4-3x^4-9x^3+9x^3+27x^2-27x^2-81x+81x+243-243)/(x+3)(x+5)
=(x^4-3x^3+9x^2-27x+81)/(x+5) -243/(x+3)(x+5)
=(-243/2)(1/(x+3)-1/(x+5)) +(x^4+5x^3-8x^3-40x^2+49x^2+245x-272x-1360+1441)/(x+5)
=(-243/2)(1/(x+3)-1/(x+5))+x^3-8x^2+49x-272+ (1441)/(x+5)
=(-243/2)(1/(x+3))+(1441+243/2)(1/(x+5))+x^3-8x^2+49x-272
y'=(243/2)(1/(x+3)^2)-(1441+243/2)(1/(x+5)^2)+3x^2-16x+49
y''=(-243)(1/(x+3)^3)+3125*(1/(x+5)^3) +6x-16
y'''=(729)(1/(x+3)^4) -9375/(x+5)^4+6
n>4
y^(n)=(-243/2)n!*(-1)^n *(1/(x+3)^(n+1)) +(3125/2)n!*(-1)^n *(1/(x+5)^(n+1)

2
f(xy)=f(x)+f(y)
∫[0,1] f(1+x)dx/(1+x^2)=∫[0,1]f(1+x)darctanx
=arctanx f(1+x)|[0,1] -∫[0,1]arctanxdf(1+x)

Answer 2

好的!