求不定积分∫1/(√x+1)
\u6c42\u4e0d\u5b9a\u79ef\u5206\u221a\ufe59x+1\ufe5a\uff0d1/\u221a\ufe59x+1\uff09+1\u5206\u6bcd\u6709\u7406\u5316
\u539f\u5f0f=\u222b[(x+1-2\u221a(x+1)+1]/(x+1-1)] dx
=\u222b[1+1/x-2\u221a(x+1)/x] dx
\u6c42\u222b\u221a(x+1)/x dx
\u4ee4a=\u221a(x+1)
x=a²-1
dx=2ada
\u222b\u221a(x+1)/x dx=\u222ba/(a²-1)*2ada
=2\u222ba²/(a²-1)da
=2\u222b(1+1/(a²-1)]da
=2\u222b[1+1/2[1/(a-1)-1/(a+1)]da
=2a+ln|(a-1)/(a+1)]+C
=2\u221a(x+1)+ln|[\u221a(x+1)-1]/[\u221a(x+1)+1]+C
\u6240\u4ee5\u539f\u5f0f=x+ln|x|-4\u221a(x+1)-2ln|[\u221a(x+1)-1]/[\u221a(x+1)+1]|+C
\u222b(1/x^2)tan(1/x)dx
=-\u222b tan(1/x) d(1/x)
=-\u222b sin(1/x)/cos(1/x) d(1/x)
=\u222b 1/cos(1/x) d(cos(1/x))
=ln|cos(1/x)| + C
\u5e0c\u671b\u53ef\u4ee5\u5e2e\u5230\u4f60\uff0c\u4e0d\u660e\u767d\u53ef\u4ee5\u8ffd\u95ee\uff0c\u5982\u679c\u89e3\u51b3\u4e86\u95ee\u9898\uff0c\u8bf7\u70b9\u4e0b\u9762\u7684"\u9009\u4e3a\u6ee1\u610f\u56de\u7b54"\u6309\u94ae\u3002
\u662f\u5426\u53ef\u4ee5\u89e3\u51b3\u60a8\u7684\u95ee\u9898\uff1f
x = (t - 1)^2
（符号 ^2 表示平方）
dx = d[(t-1)^2] = 2(t-1)dt
∫1/(√x+1)dx
= ∫2(t-1)/t dt
= 2∫dt - 2∫dt/t
= 2t - 2 ln|t| + 常数
= 2(√x + 1) - 2 ln(√x + 1) + C
把分母设为t
=∫(x+1)^(-1/2)d（x+1）=2（x+1）^（1/2）
绛旓細鎴戠殑 涓嶅畾绉垎1鍜姹 涓嶅畾绉垎1鍜嬫眰... 涓嶅畾绉垎1鍜嬫眰 灞曞紑 鎴戞潵绛 1涓洖绛 #鐑# 涓轰粈涔堢幇鍦ㄦ儏鏅枩鍓ц秺鏉ヨ秺灏戜簡? wjl371116 2017-12-08 路 鐭ラ亾鍚堜紮浜烘暀鑲茶瀹 wjl371116 鐭ラ亾鍚堜紮浜烘暀鑲茶瀹 閲囩撼鏁:15454 鑾疯禐鏁:65385 鍚慣A鎻愰棶 绉佷俊TA 鍏虫敞 灞曞紑鍏ㄩ儴 鏈洖绛旂敱鎻愰棶鑰呮帹鑽 宸茶禐杩 ...
绛旓細鎴戠殑 姹備笉瀹氱Н鍒1 3涓洖绛 #璇濋# 鎵撳伐浜哄繀鐪嬬殑鑱屽満銆庣淮鏉冦忔寚鍗!fnxnmn 2014-12-09 路 TA鑾峰緱瓒呰繃5.8涓囦釜璧 鐭ラ亾澶ф湁鍙负绛斾富 鍥炵瓟閲:1.1涓 閲囩撼鐜:14% 甯姪鐨勪汉:1浜 鎴戜篃鍘荤瓟棰樿闂釜浜洪〉 鍏虫敞 灞曞紑鍏ㄩ儴 鏇村杩介棶杩界瓟 杩界瓟 鏈洖绛旂敱鎻愰棶鑰呮帹鑽 宸茶禐杩 宸茶俯杩< 浣犲杩欎釜鍥炵瓟鐨勮瘎...
绛旓細=ln(鈭((1+胃)/(1-胃)))+C =ln|sect锛媡ant|锛婥 =lnl鈭(1+tan^2t)+tantl+c =lnl鈭(1+x^2)+xl+c
绛旓細璁＄畻杩囩▼濡備笅锛鈭 1/(1+sin^2x)dx = 鈭 [1/cos^2x]/(1/cos^2x+tan^2x)dx = 鈭 [sec^2x]/(sec^2x + tan^2x)dx = 鈭 1/(1 + 2tan^2x)dtanx = 1/鈭2 *鈭 1/(1 + (鈭2tanx)^2)d(鈭2tanx)= 1/鈭2 * arctan(鈭2tanx) + C锛圕涓哄父鏁帮級...
绛旓細缁忔祹鏁板鍥㈤槦涓轰綘瑙ｇ瓟锛屾弧鎰忚閲囩撼锛
绛旓細鈭1/(1+e^x)dx =鈭紙1+e^x-e^x锛/(1+e^x)dx =鈭1dx-鈭紙e^x锛/(1+e^x)dx =x-鈭1/(1+e^x)d锛坋^x锛=x-鈭1/(1+e^x)d锛1+e^x锛=x-ln锛1+e^x锛+C
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
绛旓細瑙ｏ細浠³鈭(x+1)=t锛屽垯x=t³-1 鈭1/[(1+³鈭(x+1)] dx =鈭玔1/[(1+t)]d(t³-1)=鈭玔2t²/[(1+t)]dt =鈭玔(2t²+2t-2t-2+2)/[(1+t)]dt =鈭玔2t-2 +2)/[(1+t)]dt =鈭(2t-2)dt+ 2鈭玔1/[(1+t)]d(1+t)=t²-...
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=ln|t+1|/3-1/6ln(t^2-t+1)+1/2arctan(t-1/2)+c 瑙ｉ噴 鏍规嵁鐗涢】-鑾卞竷灏艰尐鍏紡锛岃澶氬嚱鏁扮殑瀹氱Н鍒嗙殑璁＄畻灏卞彲浠ョ畝渚垮湴閫氳繃姹備笉瀹氱Н鍒鏉ヨ繘琛屻傝繖閲岃娉ㄦ剰涓嶅畾绉垎涓庡畾绉垎涔嬮棿鐨勫叧绯伙細瀹氱Н鍒嗘槸涓涓暟锛岃屼笉瀹氱Н鍒嗘槸涓涓〃杈惧紡锛屽畠浠粎浠呮槸鏁板涓婃湁涓涓绠楀叧绯汇備竴涓嚱鏁帮紝鍙互瀛樺湪涓嶅畾绉垎...
