大1 微積分 麻煩各位幫忙解答

1.∫(0~4)(x-1)dx/(x^2-4x+5)=∫(x-2+1)dx/[(x^2-4x+4)+1]=∫[(x-2)dx+dx]/[(x-2)^2+1]=∫[(x-2)d(x-2)/[(x-2)^2+1]+∫dx/[(x-2)^2+1]=∫ydy/(y^2+1)+∫d(x-2)/[(x-2)^2+1]......y=x-2=∫d(y^2+1)/2(y^2+1)+∫dy/(y^2+1)=0.5*ln(y^2+1)+atan(y)+c=0.5*ln[(x-2)^2+1]+atan(x-2)......x=0~4=0.5*ln(4+1)+atan(2)-0.5*ln(4+1)-atan(-2)=atan(2)+atan(2)=2*atan(2) 2.∫(x^2+2x-1)dx/(2x^3+3x^2-2x)=∫(x^2+2x-1)dx/x(2x-1)(x+2)=∫dx/2x-∫dx/10(x+2)+2∫dx/(2x-1)......Note=0.5*ln(x)-0.1*ln(x+2)+ln(2x-1)+ln(c)=ln{c(2x-1)√x/(x+2)^0.1}
Note: 部份分式(x^2+2x-1)/x(2x-1)(x+2)=a/x+b/(x+2)+c/(2x-1)f(x)=x^2+2x-1=a(x+2)(2x-1)+bx(2x-1)+cx(x+2)f(0)=-1=-2a => a=1/2f(-2)=-1=(-2)*(-5)b => b=-1/10f(1/2)=1/4=0.5*2.5c => c=2 3.∫(2x^2-x+4)dx/x(x^2+4)=∫dx/x+∫(x-1)dx/(x^2+4)......Note=ln(x)+∫xdx/(x^2+4)-∫dx/(x^2+4)=ln(x)+∫d(x^2+4)/(x^2+4)-∫d(x/2)/2[(x/2)^2+1]=ln(x)+ln(x^2+4)-∫dy/2(y^2+1)......y=x/2=ln[x(x^2+4)]-0.5*atan(y)+ln(c)=ln[cx(x^2+4)]-0.5atan(x/2) Note: 部份分式(2x^2-x+4)/x(x^2+4)=a/x+(bx+c)/(x^2+4)2x^2-x+4=a(x^2+4)+(bx+c)x=(a+b)x^2+cx+4a=> 4a=4 => a=1c=-1a+b=2 => b=2-a=1


2014-05-27 07:51:32 補充：
4.∫(0~1)dx/(x^2-4)

=∫dx/(x+2)(x-2)

=∫dx/4(x-2)-∫dx/4(x+2)

=0.25*ln(x-2)-0.25*ln(x+2)

=0.25*ln|(x-2)/(x+2)|.....x=0~1

=0.25*{ln|-1/3|-ln|-2/2|}

=0.25*[ln(1/3)-ln(1)]

=0.25*ln(1/3)

=-0.25*ln(3)

2014-06-01 18:53:15 補充：
^ 是什麼東西


ANS: 冪次方

到下面的網址看看吧

▶▶http://candy5660601.pixnet.net/blog

到下面的網址看看吧

▶▶http://candy5660601.pixnet.net/blog

這有類似的

▶▶http://qoozoo04480609.pixnet.net/blog

到下面的網址看看吧

▶▶http://candy5660601.pixnet.net/blog

參考下面的網址看看

http://phi008780520.pixnet.net/blog

參考下面的網址看看

http://phi008780520.pixnet.net/blog

第二題部分分式的c好像算錯了

2014-05-27 16:57:19 補充：
第三題積分x^2+4的地方少了1/2

第四題直接代公式就好

不會連這都懶吧