不定積分（Ａ）１、求下列不定積分２）１）dxdxx2xx21x４）2３）x2(x2)dx2dx５）23x52x６）cos2xdxdxx3xcosxsin22７）(2ex3)dx(11)xxdx８）2xx２、求下列不定積分（第一換元法）２）dx１）(32x)3dx23x3３）sintdxdt４）xlnxln(lnx)t５）６）dxdxeexcosxsinxx3x3dx８）4xcos(x2)dx７）1x1x10）94x2９）sinxdxcosdx3xdxcosxdx2x1312）211）sin2xcos3xdxtan3xsecxdx13)14)13cos2x4sin2xx3dxdx9x215)16)102arccosxdx1x2arctanxdxx(1x)17)18)13、求下列不定積分（第二換元法）1sinxdxdxx1x2１）２）４）x24x2dx,(a0)dxa2x2x３）５）７）dxdx(x21)312x６）８）dxdxx1x211x2４、求下列不定積分（分部積分法）xsinxdxarcsinxdx１）２）e2xsindxx2lnxdxx2３）４）６）x2arctanxdxx2cosxdx５）x2cos2xln2xdxdx2７）８）５、求下列不定積分（有理函數(shù)積分）x3dxx3１）2x3x23x10dx２）dxx(x21)3）（Ｂ）1、一曲線通過(guò)點(diǎn)(e2,3)，且在任一點(diǎn)處的切線斜率等于該點(diǎn)的橫坐標(biāo)的倒數(shù)，求該曲線的方程。11x23，且當(dāng)時(shí)函數(shù)值為，試求此函數(shù)。F(x)12x2、已知一個(gè)函數(shù)的導(dǎo)函數(shù)為2f(x)dxF(x)c3、證明：若，則1f(axb)dxF(axb)c,(a0)a。sinxxf(x)dx。f(x)x4、設(shè)的一個(gè)原函數(shù)為，求5、求下列不定積分xcos2dx21sin2xdx１）２）arctan11x1xxxdxdx1x2３）５）４）dx(x2a2)(x2b2)xaxx2dx６）xearctanxlnxdxdx(1x2)32x1lnx7）8）（Ｃ）求以下積分xexdxsin(2x)2sinxdxex1１）2)arctanexx5dxdx1sinxcosxsinxcosx4x3e2x３）５）４）x5xx81dxdx６）x1cosxcos222５、（１）用倍角公式：（２）注意cosxsinx0或cosxsinx0兩種情況。arctan1arccotx,1x2dxd(arccotx)1x（３）利用。（４）先分子有理化，在分開(kāi)作三角代換。（５）化為部分分式之和后積分。3（６）可令x2asinxa(ba)sin2t,則bx(ba)cos（７）可令2t。2t。（８）令1lnxt。（９）分部積分后移項(xiàng)，整理。（１０）湊earctanx后分部積分，再移項(xiàng)，整理。tanxt（１１）令。2dxx3(x2)4x2x3t2x（１２）變形為后，令，11x2t2，兩端微分得(x2)2dx2tdt1再由。（Ｃ）2u1u2xln(1u),dxdu2ue1x1）解：令，則2ln(1u2)du2uln(1u2)14uu22du所以原式2uln(1u2)4u4arctanuc2xex14ex14arctanex1c2）解：方法一：xxd()d(tan)1dx222sinx(1cosx)4x142xxxsincos32tancos222原式x1tantan212d(tan)1tan21lntanxcxx428242x2tanxt2方法二：令4sinxdx2(1cos2x)(1cosx)cosxu，然后令方法三：變形為再化成部分分式積分。1arctanexd(e2x)3）解：原式21[e2xarctanexd(ex)]e2x(1e2x)21[e2xarctanexduuu2)(12]2eu（令）x1[e2xarctanexdu1u2]du2u21e2xarctanexexarctanexc2x3x31d(x)11d(x3)1[d(x3)]3334x314x314x314）解：原式1[(x31)4d(x31)(x31)14d(x31)]334(x31)44(x31)4c21739xx3dx1d(x2x2)(x2x2)22xx42uxx2，令245）解：原式dulnu22121c2u42u21x42x21cln42x42x212sinxcosx11dx6）解：原式12sinxcosx1(sinxcosx)2dx112sinxcosxdx2sinxcosx5d(x)41(sinxcosx)1sin(x)4222dcos(x)1(sinxcosx)142221cos2(x