高等数学基础题,关于不定积分的题,第4大题,要过程,写在纸上拍照最好
2个回答
展开全部
(1) =1/6sin^6x+C
(2) =∫cos^2xd(sinx)
=∫(1-sin^2x)d(sinx)
=sinx-1/3sin^3x+C
(3) =∫xdx+2∫sin√xd√x
=1/2x^2-2cos√x+C
(4) =1/2∫e^(x^2)d(x^2)
=1/2e^(x^2)+C
(5) =-1/2∫1/√(1-x^2)d(1-x^2)
=-1/2×2√(1-x^2)+C
=-√(1-x^2)+C
(6) =1/2∫1/√[1-(x^2)^2]dx^2
=1/2arcsinx^2+C
(7) =∫(ln2+lnx)/xdx
=∫(ln2)/xdx+∫(lnx)/xdx
=ln2lnx+∫lnxd(lnx)
=ln2lnx+1/2(lnx)^2+C
(8) =∫e^x/[(e^x)^2-1]dx
=1/2∫[1/(e^x-1)-1/(e^x+1)]d(e^x)
=1/2∫1/(e^x-1)d(e^x-1)-1/2∫1/(e^x+1)d(e^x+1)
=1/2ln(e^x-1)-1/2ln(e^x+1)+C
=1/2ln[(e^x-1)/(e^x+1)]+C
(9) =∫1/arcsinxd(arcsinx)
=lnarcsinx+C
(10) =∫1/arctanxd(arctanx)
=lnarctanx+C
(2) =∫cos^2xd(sinx)
=∫(1-sin^2x)d(sinx)
=sinx-1/3sin^3x+C
(3) =∫xdx+2∫sin√xd√x
=1/2x^2-2cos√x+C
(4) =1/2∫e^(x^2)d(x^2)
=1/2e^(x^2)+C
(5) =-1/2∫1/√(1-x^2)d(1-x^2)
=-1/2×2√(1-x^2)+C
=-√(1-x^2)+C
(6) =1/2∫1/√[1-(x^2)^2]dx^2
=1/2arcsinx^2+C
(7) =∫(ln2+lnx)/xdx
=∫(ln2)/xdx+∫(lnx)/xdx
=ln2lnx+∫lnxd(lnx)
=ln2lnx+1/2(lnx)^2+C
(8) =∫e^x/[(e^x)^2-1]dx
=1/2∫[1/(e^x-1)-1/(e^x+1)]d(e^x)
=1/2∫1/(e^x-1)d(e^x-1)-1/2∫1/(e^x+1)d(e^x+1)
=1/2ln(e^x-1)-1/2ln(e^x+1)+C
=1/2ln[(e^x-1)/(e^x+1)]+C
(9) =∫1/arcsinxd(arcsinx)
=lnarcsinx+C
(10) =∫1/arctanxd(arctanx)
=lnarctanx+C
本回答被提问者采纳
已赞过
已踩过<
评论
收起
你对这个回答的评价是?
推荐律师服务:
若未解决您的问题,请您详细描述您的问题,通过百度律临进行免费专业咨询