急~求不定积分！∫(1/x+1)ln(x+1)d(x)
\u6c42\u4e0d\u5b9a\u79ef\u5206\uff1ax*ln(1+x)dx\u539f\u5f0f=1/2\u222bln(x+1)dx²
=1/2*x²ln(x+1)-1/2\u222bx²dln(x+1)
=1/2*x²ln(x+1)-1/2\u222bx²/(x+1) dx
=1/2*x²ln(x+1)-1/2\u222b(x²-1+1)/(x+1) dx
=1/2*x²ln(x+1)-1/2\u222b[x-1+1/(x+1)] dx
=1/2*x²ln(x+1)-1/4*x²+1/2x-1/2ln(x+1)+C
\u222b(dx/(1-x^2))ln[(1+x)/(1-x)]
=1/2*\u222b[1/(1+x)+1/(1-x)]*[ln(1+x)-ln(1-x)]dx
=1/2*\u222bd[ln(1+x)-ln(1-x)]
=1/2*[ln(1+x)-ln(1-x)]+C
∫1/(x+1)ln(x+1)dx = ∫ ln(x+1)dln(x+1) = (1/2)[ln(x+1) ]^2 + C
绛旓細鍥炵瓟锛氣埆1/(x+1)ln(x+1)dx = 鈭 ln(x+1)dln(x+1) = (1/2)[ln(x+1) ]^2 + C
绛旓細锛1锛鈭紙1-3x鐨2娆℃柟锛塪x =x-x^3+C (2)鈭埆a鐨3x娆℃柟dx =1/3鈭玜^3xd3x =1/(3lna)鈭玜^3x*lnad3x =a^3x/(3lna)+C (3)鈭(2鐨剎娆℃柟+x鐨2娆℃柟)dx =鈭2^xdx+鈭玿^2dx =2^x/ln2+x^3/3+C
绛旓細x+C 鍏朵腑C涓轰换鎰忓父鏁
绛旓細姹備笉瀹氱Н鍒锛1銆傗埆[(1/x²)e^(1/x)dx=-鈭玠e^(1/x)=-e^(1/x)+C 2銆傗埆xcos(2x²-1)dx=(1/4)鈭玠[sin(2x²-1)]=(1/4)sin(2x²-1)+C 3銆鈭(1-sinx)²dx=鈭(1-2sinx+sin²x)dx=x+2cosx+鈭(1/2)(1-cos2x)dx =x+2cosx+(1/...
绛旓細浠 u=ln(2x)锛屽垯 du = 1/x锛屽師寮 = 鈭1/(2u) du = 1/2 lnu + C = 1/2 ln(ln(2x))+C 浠 x=u^2锛屽垯 dx=2udu锛屽師寮 = 鈭2cosudu = 2sinu + C = 2sin鈭歺 + C
绛旓細绠鍗曟楠ゅ涓: 鈭 1/[x²鈭(x²+a²)] dx 浠=a*tany,dx=a*sec²y dy =a鈭 sec²y/[(a²*tan²y)*鈭(a²*tan²y+a²)] dy =a鈭 sec²y/[(a²*tan²y)*鈭歛²sec²y] dy =a/a³...
绛旓細1.=鈭1+sinxdx=x-cosx+C 2.=-鈭2^(1/x)d(1/x)=-2^(1/x)/ln2+C 3.=鈭玡^(x/2)鈭(1-1/e^x)dx =2鈭垰(1-1/e^x)de^(x/2)浠^(x/2)=cscu=1/sinu =2鈭玞osudcscu =-2鈭玞ot²udu =-2鈭玞sc²u-1du =2cotu+2u+C 4.=xln(x²+1)-鈭玿*...
绛旓細1 浠=t^2 鍘熷紡锛濃埆2t/(1+t)dt=2t-鈭2/(1+t)dt=2t-2ln(1+t) +c 2,鍚1涓鏍风殑棰樼洰銆3锛屽師寮忥紳鈭玜rctanxdarctanx=1/2arctan^2 x+c 4, 鍘熷紡=鈭玞ost dt^2 =鈭2tcostdt=鈭2tdsint= 2tsint-2鈭玸intdt=2tsint+2cost+c =2鈭歺 sin鈭歺 +2cos鈭歺 +c ...
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绛旓細绠鍗曞垎鏋愪竴涓嬶紝绛旀濡傚浘鎵绀
