幫解這兩題微積分問題!!!!!!

S(3sin x + 4cos x) dx

= S(3sin x) dx + S(4cos x) dx [分解兩部分積分]

= 3 S(sinx) dx + 4 S(cos x) dx [把常數抽出]

= -3cos x + 4sin x + C [代入積分公式、得出答案]

S(1-e^x)/(1+e^x) dx

= -S(e^x-1)/(e^x+1) dx [把負號抽出，使(1-e^x)變為(e^x-1)]

= -S{(e^x+1)-2}/(e^x+1) dx [把(e^x-1)分解為(e^x+1)-2]

= -[S{(e^x+1)/(e^x+1)} dx - 2 S{1/(e^x+1)} dx] [把{(e^x+1)-2}/(e^x+1) 分解為部分分式]

= -[S(1) dx - 2 S{1/(e^x+1)} dx]

= -[x-2ln(e^x+1)] [代入積分公式]

= -x + 2ln(e^x+1) + C [得出答案]

to食碗麵,,hsu: +2ln 這邊是不是有錯ㄚ!?我自己算一次都是 " - " 的
還是我有地方算錯了呢

1. Find∫(3Sin x+4Cos x) dx
Sol
∫(3Sin x+4Cos x) dx
=3∫Sin xdx+4∫Cos x dx
=－3Cosx+4Sinx+c

2.Find ∫(1－e^x)/(1+e^x)dx
Sol
(1－e^x)/(1+e^x)dx
=∫(1+e^x－2e^x)/(1+e^x)dx
=∫dx－2∫e^x/(1+e^x)dx
=∫dx－2∫de^x/(1+e^x)
=∫dx－2∫d(1+e^x)/(1+e^x)
=x－2ln(1+e^x)+c

∫ (1-e^x)/(1+e^x) dx
=∫ [e^(-x/2)-e^(x/2)]/[e^(-x/2)+e^(x/2)] dx
[提出e^(x/2)]
= -2 ∫ d[e^(-x/2)+e^(x/2)]/[e^(-x/2)+e^(x/2)] dx
=-2 ln [e^(-x/2)+e^(x/2)]

= x - 2 ln [1+e^x]

= - x + 2 ln [1+e^-x]

1. -3cosx+4sinx +c
2. S(1-e^x)/(1+e^x)dx.= S1/(1+e^x)dx - Se^x/(1+e^x)dx.
= - ln(1+e^(-x)) - ln(1+e^x) +c

應該沒解錯 有錯在講