不定积分计算代数换元法应用举例

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-289ˣ).

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-289ˣ).

=∫(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。