学帮网不定积分1/[(x+1)(x^2+1)^(1/2)]的原函数求法
来源：学生作业学帮网 编辑：学帮网 时间：2024/04/29 18:39:22
不定积分1/[(x+1)(x^2+1)^(1/2)]的原函数求法
令x=tant
则dx=sec^2 tdt
式子化为：∫1/[(tant+1)sect]* sec^2 t dt=∫dt/(sint+cost)=√2∫dt/sin(t+π/4)
由∫du/sinu=sinudu/(sinu)^2=-∫d(cosu)/(1-cos^2 u)=-0.5∫d(cosu)*[1/(1-cosu)+1/(1+cosu)]=-0.5[-ln(1-cosu)+ln(1+cosu)]=-0.5ln[(1+cosu)/(1-cosu)]=-0.5ln(1+cosu)^2/(sinu)^2=ln|sinu|/(1+cosu)
因此原式=√2ln|sin(t+π/4)|/(1+cos(t+π/4)+C,代入t即得.
令x=tant
则dx=sec^2 tdt
式子化为：∫1/[(tant+1)sect]* sec^2 t dt=∫dt/(sint+cost)=√2∫dt/sin(t+π/4)
由∫du/sinu=sinudu/(sinu)^2=-∫d(cosu)/(1-cos^2 u)=-0.5∫d(cosu)*[1/(1-cosu)+1/(1+cosu)]=-0.5[-ln(1-co...
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令x=tant
则dx=sec^2 tdt
式子化为：∫1/[(tant+1)sect]* sec^2 t dt=∫dt/(sint+cost)=√2∫dt/sin(t+π/4)
由∫du/sinu=sinudu/(sinu)^2=-∫d(cosu)/(1-cos^2 u)=-0.5∫d(cosu)*[1/(1-cosu)+1/(1+cosu)]=-0.5[-ln(1-cosu)+ln(1+cosu)]=-0.5ln[(1+cosu)/(1-cosu)]=-0.5ln(1+cosu)^2/(sinu)^2=ln|sinu|/(1+cosu)
因此原式=√2ln|sin(t+π/4)|/(1+cos(t+π/4)+C, 代入t即得。
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