利用分部积分法,求不定积分。 利用分部积分法求下列不定积分
\u5229\u7528\u4e0d\u5b9a\u79ef\u5206\u7684\u5206\u90e8\u79ef\u5206\u6cd5\u6c42\u4e0d\u5b9a\u79ef\u5206\u222bxsinx/cos³xdx
\u56e0\u4e3a\uff1a(1/cosx)'=(sinx/cos²x)
\u539f\u5f0f=\u222bx/cosxd(1/cosx) \u5206\u90e8\u79ef\u5206
=x/cos²x-\u222b1/cosxd(x/cosx)
=x/cos²x-\u222b1/cosx*(cosx+xsinx/cos²x)dx
=x/cos²x-\u222b1/cos²xdx-\u222bxsinx/cos³xdx
\u4ee4\u222bxsinx/cos³xdx=F
\u5219F=x/cos²x-\u222b1/cos²xdx-F
2F=x/cos²x-\u222b1/cos²xdx=x/cos²x-\u222bsec²xdx
=x/cos²x-tanx+C
\u6545\u539f\u79ef\u5206=(x/cos²x-tanx)/2+C
=2∫u-2+2/(u+1)du
=u²-4u+4ln(u+1)+C,代入u=√(x+1)
绛旓細=xlnx-x+C
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绛旓細鍒嗘绉垎娉 鍘熷紡=xarctan鈭歺-鈭玿darctan鈭歺 =xarctan鈭歺-鈭玿/(1+x)dx =xarctan鈭歺-鈭(x+1-1)/(1+x)dx =xarctan鈭歺-鈭玔1-1/(1+x)]dx =xarctan鈭歺-x+ln(1+x)+C
绛旓細绉垎杩囩▼涓 浠 = sin胃锛屽垯dx = cos胃 d胃 鈭垰(1-x²)dx =鈭垰(1-sin²胃)(cos胃 d胃)=鈭玞os²胃d胃 =鈭(1+cos2胃)/2d胃 =胃/2+(sin2胃)/4+C =(arcsinx)/2+(sin胃cos胃)/2 + C =(arcsinx)/2+(x鈭(1 - x²))/2+C =(1/2)[arcsinx...
绛旓細=鈭(u-1)/(u+1)d(u²-1)=2鈭玼-2+2/(u+1)du =u²-4u+4ln(u+1)+C锛屼唬鍏=鈭(x+1)
绛旓細銆
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