几道微积分的题,比较迷惑恳请大神解答!!
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1.∫(x+1)/(x²+4x+4) dx
=∫(x+2–1)/(x²+4x+4) dx
=∫(x+2)/(x²+4x+4)dx–∫dx/(x²+4x+4)
=∫dx/(x+2)–∫dx/(x+2)²
=ln|x+2|+1/(x+2)+C
2.∫(2x+3)/(x²+2x+2) dx
=∫(2x+2+1)/(x²+2x+2) dx
=∫(2x+2)/(x²+2x+2) dx+∫dx/(x²+2x+2)
=∫1/(x²+2x+2) d(x²+2x+2) +∫dx/[(x+1)²+1]
=ln(x²+2x+2)+arctan(x+1)+C
3.∫(x–2)dx/[(x+2)(x²–x+1)]
=∫[–4/7(x²–x+1)+(4/7 x–5/7)(x+2)]dx/[(x+2)(x²–x+1)]
=–4/7 ∫dx/(x+2)+1/7 ∫(4x–5)dx/(x²–x+1)
=–4/7∫dx/(x+2)+1/7 ∫(4x–2–3)dx/(x²–x+1)
=–4/7∫dx/(x+2)+2/7∫(2x–1)dx/(x²–x+1)–3/7∫dx/[(x–1/2)²+(∨3/2)²]
=–4/7 ln|x+2|+2/7 ln(x²–x+1)–2∨3/7 arctan[(2x–1)/∨3] +C
附:
令(x–2)/[(x+2)(x²–x+1)]=a/(x+2)+(bx+c)/(x²–x+1)
则ax²–ax+a+bx²+cx+2bx+2c=x–2
a+b=0,2b+c–a=0,a+2c=–2
c=–5/7
a=–4/7
b=4/7
所以(x–2)/[(x+2)(x²–x+1)]
=–4/7 · 1/(x+2)+(4/7 x–5/7)/(x²–x+1)
函数f(x)求导后会出现分式的,最常见的几种情况
①f(x)=m·(g(x))^a + C (a为负整数),
← → f'(x)=a·m·g'(x)/[g(x)]^(1–a)
②f(x)=m·ln|g(x)| + C, ← →f'(x)=[m·g'(x)]/g(x)
③f(x)=m·arctan[g(x)]+C,←→ f'(x)=m·g'(x) /[1+(g(x))²]
(m为常数)
在求被积函数是分式形式的不定积分的时候,可以将被积函数试着进行上面几种形式进行拆分。
=∫(x+2–1)/(x²+4x+4) dx
=∫(x+2)/(x²+4x+4)dx–∫dx/(x²+4x+4)
=∫dx/(x+2)–∫dx/(x+2)²
=ln|x+2|+1/(x+2)+C
2.∫(2x+3)/(x²+2x+2) dx
=∫(2x+2+1)/(x²+2x+2) dx
=∫(2x+2)/(x²+2x+2) dx+∫dx/(x²+2x+2)
=∫1/(x²+2x+2) d(x²+2x+2) +∫dx/[(x+1)²+1]
=ln(x²+2x+2)+arctan(x+1)+C
3.∫(x–2)dx/[(x+2)(x²–x+1)]
=∫[–4/7(x²–x+1)+(4/7 x–5/7)(x+2)]dx/[(x+2)(x²–x+1)]
=–4/7 ∫dx/(x+2)+1/7 ∫(4x–5)dx/(x²–x+1)
=–4/7∫dx/(x+2)+1/7 ∫(4x–2–3)dx/(x²–x+1)
=–4/7∫dx/(x+2)+2/7∫(2x–1)dx/(x²–x+1)–3/7∫dx/[(x–1/2)²+(∨3/2)²]
=–4/7 ln|x+2|+2/7 ln(x²–x+1)–2∨3/7 arctan[(2x–1)/∨3] +C
附:
令(x–2)/[(x+2)(x²–x+1)]=a/(x+2)+(bx+c)/(x²–x+1)
则ax²–ax+a+bx²+cx+2bx+2c=x–2
a+b=0,2b+c–a=0,a+2c=–2
c=–5/7
a=–4/7
b=4/7
所以(x–2)/[(x+2)(x²–x+1)]
=–4/7 · 1/(x+2)+(4/7 x–5/7)/(x²–x+1)
函数f(x)求导后会出现分式的,最常见的几种情况
①f(x)=m·(g(x))^a + C (a为负整数),
← → f'(x)=a·m·g'(x)/[g(x)]^(1–a)
②f(x)=m·ln|g(x)| + C, ← →f'(x)=[m·g'(x)]/g(x)
③f(x)=m·arctan[g(x)]+C,←→ f'(x)=m·g'(x) /[1+(g(x))²]
(m为常数)
在求被积函数是分式形式的不定积分的时候,可以将被积函数试着进行上面几种形式进行拆分。
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