# 請問一些微積分的問題

總共有4題

但是因為顯這裡是不出來

所以我把問題貼到我的相簿上

雖然會造成您的不便

但還煩請您耐心的幫我解決疑惑

謝謝

http://www.wretch.cc/album/show.php?i=dg456trgv&b=...

請詳細解說為什麼上式=下式

還有第3題 我不知道哪個是正確答案

所以且挑出正確答案並解釋為什麼上式=下式

拜託大大了 謝謝

### 2 Answers

- 1 decade agoFavorite Answer
1.

Consider (secx)' = secx*tanx and (tanx)' = secx^2

Ans =∫secx(secx + tanx) / (secx + tanx) dx

= ∫(secx^2 + secxtanx) / (secx + tanx) dx

= ln |secx + tanx| + C

2.

Consider [√(x^2 + 2x + 26)]' = (2x + 2) / 2√(x^2 + 2x + 26)

Ans=∫d(x^2 + 2x + 26) / √(x^2 + 2x + 26) dx

= 2√(x^2 + 2x + 26) + C

( or you can let t = x^2 + 2x + 26, dt = (2x + 2)dx )

3.

Consider [ln(x^2 + 1)]' = 2x / (x^2 + 1)

Ans = (1/2)∫d(x^2 + 1) / (x^2 + 1) dx

= (1/2) ln |x^2 + 1|

= (1/2) ln (x^2 + 1) + C

since both x^2 and 1 are positive, so we have ln |x^2 + 1| = ln(x^2 + 1)

( or you can let t = x^2 + 1, dt = 2xdx )

4.

Consider [ln(x^2 + 2)]' = 2x/(x^2 + 2) and [1/(x^2 + 2)]' = -2x / [(x^2 + 2)^2]

Ans=∫(1/2) d(x^2 + 2)/(x^2 + 2) dx - 5∫(1/2)d(x^2 + 2)/(x^2 + 2)^2 dx

= -(1/2)ln|x^2 + 2| + (5/2)[1/(x^2 + 2)]+C

( or you can let t = x^2 + 2, dt = 2xdx )

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