求解高数不定积分 高数不定积分求解
\u9ad8\u6570\uff0c\u4e0d\u5b9a\u79ef\u5206\u6c42\u89e3\u6bd4\u8f83\u590d\u6742\uff0c\u5982\u56fe\u6240\u793a\u3002
\u5206\u90e8\u79ef\u5206\u6cd5\u9006\u5e94\u7528
\u222b1/(u²+a²)du
=u/(u²+a²)-\u222bu(-2u/(u²+a²)²)du
=u/(u²+a²)+2\u222b(1/(u²+a²)-a²/(u²+a²)²)du
\u22342a²\u222b1/(u²+a²)²du=u/(u²+a²)+\u222b1/(u²+a²)du
a<0 同理 可等到积分=负四分之派
x = asinθ、dx = acosθ dθ
∫[0→a] dx/[x + √(a² - x²)]
= ∫[0→π/2] acosθ/[asinθ + acosθ] dθ
= (1/2)∫[0→π/2] 2cosθ/[sinθ + cosθ] dθ
= (1/2)∫[0→π/2] [(sinθ + cosθ) - (sinθ - cosθ)]/(sinθ + cosθ) dθ
= (1/2)∫[0→π/2] dθ - (1/2)∫[0→π/2] d(- cosθ - sinθ)/(sinθ + cosθ)
= θ/2 |[0→π/2] + (1/2)∫ d(sinθ + cosθ)/(sinθ + cosθ)
= π/4 + (1/2)ln[sinθ + cosθ] |[0→π/2]
= π/4 + (1/2){ln(1 + 0) - ln(0 + 1)}
= π/4
先通分，再计算，我不太方便做，你看看。
绛旓細鏂规硶濡備笅锛岃浣滃弬鑰冿細
绛旓細鍦ㄥ井绉垎涓紝鍑芥暟鐨涓嶅畾绉垎鏄竴涓〃杈惧紡锛屽畾绉垎鏄竴涓暟銆傦紝
绛旓細瑙ｏ細鍒嗕韩涓绉嶈В娉曪紝鍒╃敤鏂圭▼鑱旂珛姹傝В銆傗埖(sinx)^4+(cosx)^4=1-2(sinxcosx)^2=1-(1/2)(sin2x)^2=(3/4)+(1/4)cos4x锛(sinx)^4-(cosx)^4=(sinx)^2-(cosx)^2=-cos2x锛屸埓璁綢1=鈭(sinx)^4dx锛孖2=鈭(cosx)^4dx锛屽垯I1+I2=鈭玔3/4+(1/4)cos4x]dx=3x/4+(1/16)sin...
绛旓細13棰橈細f(x)鍘熷嚱鏁涓簒lnx锛屽垯f(x)=(xlnx)'=lnx+1 f'(x)=1/x 鈭玿f'(x)dx= 鈭玠x=x+C 14棰橈細鍘熷紡= 鈭(x^2+1-1)arctanx/(x^2+1)dx= 鈭玜rctanxdx- 鈭玜rctanx/(1+x^2)dx=x*arctanx- 鈭玿darctanx- 鈭玜rctanxd(arctanx)=x*arctanx- (1/2)鈭1/(1+x^2...
绛旓細鈭玿.e^(-x) dx =-鈭玿de^(-x)=-x.e^(-x) +鈭玡^(-x) dx =-x.e^(-x) -e^(-x) +C (2)鈭玿lnx dx =(1/2) 鈭玪nx dx^2 =(1/2)x^2.lnx -(1/2) 鈭玿 dx =(1/2)x^2.lnx -(1/4)x^2 +C (3)鈭玿^2 arctanx dx =(1/3)鈭玜rctanx dx^3 =...
绛旓細1-x^2)^(1/2)-2/3x+1/3arccosx*(1-x^2)^(3/2)-1/9x^3+C 2.sinxcosx/锛坰inx+cosx锛=sinxcosx/(2^1/2*sin(x+pi/4)),浠=pi/4+x,鍘熷紡鍖栫畝涓 1/(2*2^1/2)*(2(sinu)^2-1)/(sinu),鎵浠绉垎灏卞寲涓(1/2^1/2)*sinu-1/(2*2^1/2)*cscu.姹傝В鍗冲彲銆
绛旓細绗竴棰橈細浠锛濓紙3/2锛塻inu锛屽垯锛歴inu锛2x/3锛寀锛漚rcsin锛2x/3锛夛紝dx锛濓紙3/2锛塩osudu銆傗埓鈭蓟锛1锛峹锛/鈭氾紙9锛4x^2锛夛冀dx 锛濓紙3/2锛夆埆锝涳蓟1锛嶏紙3/2锛塻inu锛/鈭氾蓟9锛9锛坰inu锛塣2锛斤綕cosudu 锛濓紙1/4锛夆埆锛2锛峴inu锛塪u 锛濓紙1/2锛夆埆du锛嶏紙1/4锛夆埆sinudu 锛濓紙1/2锛塽锛嬶紙1...
绛旓細1銆佸垎姣嶆槸鏃犳硶鎻愬彇鍏洜寮忕殑澶氶」寮忥紝灏ゅ叾鍒嗗瓙鍒嗘瘝鏈楂樻鐩哥瓑锛岄鍏堟妸鍒嗗瓙绠鍖栵紝鐩殑鏄妸鍒嗗瓙闄嶆锛屽垎瑙ｅ嚭甯告暟椤癸紝鐒跺悗鍓╀笅閮ㄥ垎鍑戝父瑙佽绉嚱鏁板舰寮忥紝鏈鍒嗘瘝鏄簩娆★紝鎵浠ュ墿涓嬫牴鎹(arctanx)'=1/(1+x²锛夊噾褰㈠紡鍗冲彲銆2銆佸垎姣嶆槸澶氶」寮忕浉涔橈紝瑁傞」锛屽彉鎴愬拰鐨勫舰寮忥紝鐒跺悗杞寲鎴愬父瑙佽绉嚱鏁板舰寮姹傝В銆傛湰...
绛旓細涓嶈兘鍑戝井鍒嗐傚厛鎹㈠厓锛歵=鈭(e^x+1)寰楀埌锛歺=ln(t²-1)鈭磀x=2t/(t²-1)路dt 鍘熷紡=鈭玪n(t²-1)路(t²-1)/t路2t/(t²-1)路dt =鈭玪n(t²-1)路2dt =2t路ln(t²-1)-2鈭玹路2t/(t²-1)路dt =2t路ln(t²-1)-鈭4t&#...
绛旓細涓妤煎彲鎯滄病鏈夎В瀹岋紝浜屾ゼ鍙儨杩愮畻鍑洪敊銆绉垎鐨勮缁嗚繃绋嬶紝璇峰弬鐪嬪浘鐗囥傜偣鍑绘斁澶э紝鍐嶇偣鍑诲啀鏀惧ぇ銆
