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热心网友 时间:2024-11-01 23:39
不定积分计算代数换元法应用举例
例题1:∫dx/[3+³√(85x+38)].
思路:变三次立方根无理
式为有理式,变量替换t=³√(85x+38)。
解:设t=³√(85x+38),则85x+38=t³,85dx=3t²dt;
∴∫dx/[3+³√(85x+38)]
=(1/85)*∫85dx/(3+t),
=(1/85)∫3*t²dt/(3+t),
=(3/85)∫[(t-3)(t+3)+3²]dt/(3+t),
=(3/85)*[∫(t-3)dt+3²∫dt/(t+3)],
=(3/85)*[1/2*t²-3t+3²*ln|t+3|+c],
=(3/170)*t-(9/170)t+(27/170)*ln|t+3|+C,
=(3/170)*³√(85x+38)²-(9/170)* ³√(85x+38)+(27/170)*ln|³√(85x+38)+3|+C.
例题2:∫[95+24(√x)³]dx/(3+√x).
思路:变平方根无理式为有理式,变量替换t=√x.
解:设t=√x,则x=t²,dx=2tdt;
∴∫[95+24(√x)³]dx/(3+√x),
=2∫(95+24 t³)tdt/(3+t),
=2∫(24*t³-24*3*t²+24 *3²*t-553)dt
+2*553∫dt/(t+3),
=(1/2)*24*t⁴-(2/3)*24*3*t³+24*3²*
t²-2*553t+2*553*ln|t+3|+C,
=(1/2)*24*x²-(2/3)*24*3*√x³+24*3²*x-2*553*√x
+2*553*ln(√x+3)+C.
例题3:∫[√(66x+7)-23]dx/[23+√(66x+7)].
思路:变根式无理式为有理式,变量替换t=√(66x+7).
解:设t=√(66x+7),则66x+7=t²,66dx=2tdt;
∴∫[√(66x+7)-23]dx/[23+√(66x+7)]
=(1/66)∫(t-23)*2tdt/(23+t),
=(1/66)[∫(2t-4*23)dt+(2/33)*23²∫dt/(t+23)],
=(1/66)(t²-4*23t)+(2/33)*23²*ln|t+23|+C,
=(1/66)( 66x+7-4*23*√(66x+7)+
(2/33)*23²*ln[(√(66x+7)+23)]+C,
例题4:∫x√(79-21x)dx.
思路:变根式无理式√(79-21x)为有理式,变量替换t=√(79-21x).
解:设t=√(79-21x),则t²=79-21x,即:x=(1/21)(79-t²),
此时有:dx=-(1/21)*2tdt;
∴∫x√(79-21x)dx
=∫(1/21)(79-t²)*t*d[(1/21)(79-t²)],
=∫(1/21)(79-t²)*t*[-(1/21)]*2tdt,
=-2*(1/21²)∫(79-t²)*t²dt,
=-2/21²*∫(79t²-t⁴)dt,
=-2/21²*[(2/3*79*t³-(1/5)*t⁵)]+C,
=-316/(3*21²)*t³+2/(5*21²)t⁵+C,
=-316/(3*21²)*√(79-21x)³+2/(5*21²)*√(79-21x)⁵+C,
例题5:∫(14ˣ*17ˣ)dx/(196ˣ-2ˣ).
思路:将被积函数进行变形,再进行变量替换,本题变量替换t=(14/17)ˣ.
解:设t=(14/17)ˣ,则dt=t*ln(14/17)dx,
∫(14ˣ*17ˣ)dx/(196ˣ-2ˣ).
=∫(14/17)ˣdx/[1-(14/17)²ˣ],
=∫t*1/[t*ln(14/17)]dt/(1-t²),
=1/ln(17/14)*∫dt/(1-t²),
=1/(ln14-ln17)*[∫dt/(t-1)-∫dt/(t+1)],
=1/(2ln14-2ln17)*ln|(t-1)/(t+1)|+C,
=1/(2ln14-2ln17)*ln|[(14/17)ˣ-1]/[(14/17)ˣ+1]|+C,
=1/(2ln14-2ln17)*ln|(17ˣ-14ˣ)/(17ˣ+14ˣ)|+C.
例题6:∫dx/³√[(77x+73)²*(77x-73)⁴].
思路:代数式换元法,本题变量替换t=(77x+73)/(77x-73).
解:设t=(77x+73)/(77x-73),有:
-2*77*73dx/(77x-73)²=dt,
即:dx=-(77x-73)²*dt/2*77*73.
代入积分函数有:∫dx/³√[(77x+73)²*(77x-73)⁴],
=∫dx/{³√[(77x+73)/(77x-73)]²*(77x-73)²] },
=[-1/(2*77*73)]∫(77x-73)²*dt/[³√t²*(77x-73)²] ,
=[-1/(2*77*73)]*∫dt/³√t² ,
=[-3/(2*77*73)]* ³√t+C,
=-3/(11242*1²)*³√t+C,
=-3/(11242*1²)*³√[(77x+73)/(77x-73)]+C。