精选积分习题3.pdf

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1、习题6.3⒈求下列不定积分：dx23x+⑴∫；⑵∫dx；()xx−+11()2()xx22−+11()xdxdx⑶∫；⑷∫；()xx++12()23(x+3)()xx2++44(xx2++45)23dx⑸∫x3+1dx；⑹∫x42++x1；xx4+54+x3+1⑺∫xx2+54+dx；⑻∫x3+−56xdx；x2dx⑼∫1−x4dx；⑽∫x4+1；dxx2+1⑾∫；⑿∫dx；()xx22++11(x+)xx()3−1x2+21−x7⒀∫dx；⒁∫dx；()xx22++1xx()1+7x9x31n−⒂∫()xx10++2252dx；⒃∫(

2、)x22n+1dx。dx1⎛11⎞解（1）∫=⎜−⎟dx()xx−11(+)2∫⎜2⎟2⎝(x−1)(x+1)(x+1)⎠11x−1=ln++C。41xx++2(1)2x+3（2）dx∫22(x−1)(x+1)2x+3Ax+BCx+D设=+，则2222(x−1)(x+1)x−1x+122(Ax+B)(x+1)+(Cx+D)(x−1)≡2x+3，于是186⎧A+C=0⎪⎪B+D=0⎨，⎪A−C=2⎪⎩B−D=333解得A=1,C=−1,B=,D=−。所以222x+3⎛xx⎞3⎛11⎞dx=⎜−⎟dx+⎜−⎟dx∫22∫22∫22(x−1)

3、(x+1)⎝x−1x+1⎠2⎝x−1x+1⎠211xx−−313=+lnln−arctanx+C。221xx++412xdx（3）∫()xx++12()23(x+3)x设23(x+1)(x+2)(x+3)ABCDEF=+++++，则223x+1x+2(x+2)x+3(x+3)(x+3)2333A(x+2)(x+3)+B(x+1)(x+2)(x+3)+C(x+1)(x+3)2222+D(x+1)(x+2)(x+3)+E(x+1)(x+2)(x+3)+F(x+1)(x+2)≡x。13令x=−1，得到A=−；令x=−2，得到C=2；令x=−3

4、，得到F=；82再比较等式两边54x、x的系数与常数项，得到⎧AB++D=0⎪⎨13AB++12C+11D+E=0。⎪⎩108AB++5427C+36D+12E+4F=0141133于是解得A=−,B=−5,C=2,D=,E=,F=，即8842x23(x+1)(x+2)(x+3)15412133=−−++++。2238(x+1)x+28(x+3)(x+2)4(x+3)2(x+3)所以187xdx∫()xx++12()23(x+3)411(x+3)2133=−ln−−+C。4028(xx++1)(2)x+24(x+3)4(x+3)dx（4

5、）∫222(x+4x+4)(x+4x+5)111=−2222222(x+4x+4)(x+4x+5)(x+4x+4)(x+4x+5)(x+4x+5)111=−−，2222x+4x+4x+4x+5(x+4x+5)所以dx1d(x+2)=−−arctan(x+2)−∫222∫22(x+4x+4)(x+4x+5)x+2[1+(x+2)]12x+3=−−−arctan(x+2)+C。2xx++22(4x+5)23（5）∫dxx3+12⎛1x−2⎞1d(x−x+1)3dx=⎜−⎟dx=lnx+1−+∫∫22∫2⎝x+1x−x+1⎠2x−x+12x−

6、x+1122x−1=+lnx1−ln(xx−+1)+3arctan+C。2333dx1(x+1)−(x−1)（6）解一：∫x42+x+1=∫22dx2(x+x+1)(x−x+1)1(x+1)dx1(x−1)dx=−∫2∫22x+x+12x−x+1221d(x+x+1)1dx1d(x−x+1)1dx=+−+∫2∫2∫2∫24x+x+14x+x+14x−x+14x−x+1211xx++12x+12x−1=+ln[arctan+arctan]+C。241xx−+233318822dx1(1+x)dx1(1−x)dx解二：=+∫42∫42∫42

7、x+x+12x+x+12x+x+1−−111xx−1x+x+1=+arctanln+C−123341xx+−2211xx++1x−1=+lnarctan+C。241xx−+233x211xx++13x注：本题的答案也可以写成ln+arctan+C。2241xx−+231−x4x+5x+4（7）dx∫2x+5x+44x+5x+4280=x−5x+21−，2x+5x+4x+4所以4x+5x+41532dx=−xxx+21−80lnx+4+C。∫2x+5x+4323x+1（8）dx∫3x+5x−63x+15x−711x+22=1−=1+−，3

8、22x+5x−6(x−1)(x+x+6)4(x−1)4x+x+6所以32xx+−11(1)432x+1dx=+xln−arctan+C。∫32xx+−568x+x+642323x211⎛⎞111+x1（9）