求不定积分1
\u6c42\u4e0d\u5b9a\u79ef\u5206\u222b1/xdx\u7b54\uff1a
\u56e0\u4e3a\u79ef\u5206\u51fd\u6570y=f(x)=1/x\u662f\u53cd\u6bd4\u4f8b\u51fd\u6570\uff0c\u5b58\u5728\u4e24\u652f
\u6240\u4ee5\uff1ax0\u90fd\u8981\u8003\u8651
x>0\u65f6\u79ef\u5206\u5f97\uff1alnx+C
x<0\u65f6\uff1a
\u222b 1/x dx=\u222b 1/(-x) d(-x)=ln(-x)+C
\u7efc\u4e0a\u6240\u8ff0\uff0c\u222b1/x dx=ln|x|+C
x<0\u65f6\uff0cln(-x)\u7684\u5bfc\u6570\u4e5f\u662f1/x
\u7b54\uff1a
\u222b 1/[x(1+x^2)] dx
=\u222b 1/x - x/(1+x^2) dx
=lnx-(1/2)ln(1+x^2)+C
x^2= -(1-x^2) + 1
∫x^2/√(1-x^2) dx
=-∫ √(1-x^2) dx + ∫dx/√(1-x^2)
let
x= siny
dx = cosy dy
∫x^2/√(1-x^2) dx
=-∫ √(1-x^2) dx + ∫dx/√(1-x^2)
=-∫ (cosy)^2 dy + ∫ dy
= -(1/2)∫ (1+cos2y) dy + y
= -(1/2)[y+(1/2)sin(2y)) + y + C
= (1/2)( y - (1/2)sin2y) + C
=(1/2)[ arcsinx - x√(1-x^2) ] + C
绛旓細鈭 1/[x(x^10+1)] dx = 鈭 [(x^10+1)-x^10]/[x(x^10+1)] dx = 鈭 1/x dx - 鈭 x^9/(x^10+1) dx = 鈭 1/x dx - (1/10)鈭 d(x^10+1)/(x^10+1)= ln|x| - (1/10)ln|x^10 + 1| + C 杩炵画鍑芥暟锛屼竴瀹氬瓨鍦ㄥ畾绉垎鍜涓嶅畾绉垎锛涜嫢鍦ㄦ湁闄愬尯闂碵a,b]涓...
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绛旓細瀹為檯涓婏紝鍐欏埌鍊掓暟绗簩姝ュ氨鍙互浜嗐備互涓婏紝璇烽噰绾炽
绛旓細绛旓細鍥犱负绉垎鍑芥暟y=f(x)=1/x鏄弽姣斾緥鍑芥暟锛屽瓨鍦ㄤ袱鏀 鎵浠ワ細x<0鍜寈>0閮借鑰冭檻 x>0鏃剁Н鍒嗗緱锛歭nx+C x<0鏃讹細鈭 1/x dx=鈭 1/(-x) d(-x)=ln(-x)+C 缁间笂鎵杩帮紝鈭1/x dx=ln|x|+C x<0鏃讹紝ln(-x)鐨勫鏁颁篃鏄1/x
绛旓細(1)鈭玿²e^(-2x)dx =(-½)鈭玿²d[e^(-2x)]=(-½)x²路e^(-2x) +½鈭玔e^(-2x)]d(x²)=(-½)x²路e^(-2x) +½鈭玔2x路e^(-2x)]dx =(-½)x²路e^(-2x) +鈭玔x路e^(-2x)]dx =(-½)x...
绛旓細1/(sinx+cosx)鐨涓嶅畾绉垎鍏蜂綋鍥炵瓟濡備笅锛氣埆1/(sinx+cosx) dx =鈭1/[鈭2(sinxcos蟺/4+sin蟺/4路cosx)]dx =鈭1/[鈭2sin(x+蟺/4)] dx =鈭2/2 鈭玞sc(x+蟺/4) d(x+蟺/4)=鈭2/2 ln|csc(x+蟺/4)-cot(x+蟺/4)|+C 涓嶅畾绉垎鐨勫叕寮忥細1銆佲埆adx=ax+C锛宎鍜孋閮芥槸甯告暟 2銆...
绛旓細鍘熷紡=鈭玜secz*asec²z dz =鈭玸ecz dtanz,a²鍏堢渷鐣 =secztanz - 鈭玹anz dsecz =secztanz - 鈭玹anz(secztanz) dz =secztanz - 鈭玸ec³z dz + 鈭玸ecz dz 鈭2鈭玸ec³z dz = secztanz + ln|secz + tanz| 鈭粹埆sec³z dz = (1/2)secztanz + (...
绛旓細1.=鈭紙1锛2x锛1/8dx=1/2鈭紙1锛2x锛1/8d锛1锛2x锛=1/2脳锛1锛2x锛1/8锛1/锛1/8锛1锛=4/9锛1锛2x锛9/8
绛旓細璁3x+4=t x=锛坱-4锛/3 鍘熷紡=鈭紙t-4锛/9t^2dt =1/9鈭1/t-4/t^2dt =1/9ln|t|+4/(9t)+C 缁撴灉涓1/9ln|3x+4|+4/9锛3x+4锛+C
绛旓細鍋囪鏄1/sinx鐨绉垎 鈭1/sinx dx =鈭1/[2sin(x/2)cos(x/2)] dx,涓ゅ嶈鍏紡 =鈭1/[sin(x/2)cos(x/2)] d(x/2)=鈭1/tan(x/2)*sec²(x/2) d(x/2)=鈭1/tan(x/2) d[tan(x/2)], [娉ㄢ埆sec²(x/2)d(x/2)=tan(x/2)+C]=ln|tan(x/2)|+C, ...
