不定积分x^2乘根号下x/1-x
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不定积分x^2乘根号下x/1-x
表达式不够明确,可能被理解为两种情形:x^2(√x)/(1-x) 或 (x^2)*√[x/(1-x)];
如是第一种情形积分:设t=√x,则dx=2tdt;
∫[x^2(√x)/(1-x)]dx=∫[2t^6/(1-t^2]dt=-2∫t^4dt-2∫t^2dt-2∫dt+2∫[dt/(1-t^2)
=-(2/5)t^5-(2/3)t^3-2t+(1/2)ln[(1+t)/(1-t)]+C
=-(2/5)x^2√x-(2/3)x√x-2√x+(1/2)ln|(1+√x)/(1-√x)|+C;
如是第二种情形积分,有些麻烦:设t=√[x/(1-x)],x=t^2/(1+t^2),dx=2tdt/(1+t^2)^2;
∫x^2*√[x/(1-x)]dx=∫[t^2/(1+t^2)]^2*t*2tdt/(1+t^2)^2=∫[2t^6/(1+t^2)^4]dt;
再设tan u=t,则dt=du/(cosu)^2;
原积分=∫[2(tanu)^6/(1+(tanu)^2)^4] du/(cosu)^2=∫2(sinu)^6du=(1/4)∫(1-cos2u)^3 du
=(1/4)∫[1-3cos2u+3(cos2u)^2-(cos2u)^3]du=u/4-(3/8)sin(2u)+[3u/2+(3/8)sin(4u)]-[(sin2u)/2-(sin2u)^3/6]
=7u/4-(7/8)sin2u+(3/8)sin(4u)+(sin2u)^3/6+C
将u=arctan√[x/(1-x)]代入上式即得最后结果;
如是第一种情形积分:设t=√x,则dx=2tdt;
∫[x^2(√x)/(1-x)]dx=∫[2t^6/(1-t^2]dt=-2∫t^4dt-2∫t^2dt-2∫dt+2∫[dt/(1-t^2)
=-(2/5)t^5-(2/3)t^3-2t+(1/2)ln[(1+t)/(1-t)]+C
=-(2/5)x^2√x-(2/3)x√x-2√x+(1/2)ln|(1+√x)/(1-√x)|+C;
如是第二种情形积分,有些麻烦:设t=√[x/(1-x)],x=t^2/(1+t^2),dx=2tdt/(1+t^2)^2;
∫x^2*√[x/(1-x)]dx=∫[t^2/(1+t^2)]^2*t*2tdt/(1+t^2)^2=∫[2t^6/(1+t^2)^4]dt;
再设tan u=t,则dt=du/(cosu)^2;
原积分=∫[2(tanu)^6/(1+(tanu)^2)^4] du/(cosu)^2=∫2(sinu)^6du=(1/4)∫(1-cos2u)^3 du
=(1/4)∫[1-3cos2u+3(cos2u)^2-(cos2u)^3]du=u/4-(3/8)sin(2u)+[3u/2+(3/8)sin(4u)]-[(sin2u)/2-(sin2u)^3/6]
=7u/4-(7/8)sin2u+(3/8)sin(4u)+(sin2u)^3/6+C
将u=arctan√[x/(1-x)]代入上式即得最后结果;
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