prss.net
当前位置:首页 >> 求不定积分∫xDx/√(x-3)和∫√(x^2-9)Dx/x >>

求不定积分∫xDx/√(x-3)和∫√(x^2-9)Dx/x

如图

令t=√(x^2-9), t^2=x^2-9, 2tdt=2xdx tdt=xdx 积分号下:√(x^2-9)dx/x =√(x^2-9) xdx/x^2 (分子分母同乘以x) =t *tdt/(t^2+9) =t^2dt/(t^2+9) =[1-9/(t^2+9)]dt ∫[1-9/(t^2+9)]dt=t-3arctan(t/3)+C=√(x^2-9)-3arctan[√(x^2-9)/3]+C 希望对你能...

我想你的题应该是这样吧 ∫ x³/(9+x²) dx =(1/2)∫ x²/(9+x²) d(x²) =(1/2)∫ (x²+9-9)/(9+x²) d(x²) =(1/2)∫ 1 d(x²) - (9/2)∫ 1/(9+x²) d(x²) =(1/2)x² - (9/2)ln(x²+9) + C ...

无法正常回答

如图

大概是这样

∫x^3/(9+x^2)dx =1/2∫x^2/(9+x^2)dx^2 (x^2=t) =1/2∫t/(9+t)dt =1/2∫(t+9-9)/(9+t)dt =1/2∫[1-9/(9+t)]dt =1/2t-9/2ln(9+t)+C =1/2x^2-9/2ln(9+x^2)+C

(x^2 - 9)^(1/2) - log(((x^2 - 9)^(1/2) + 3*i)/x)*3*i

记t=√(x+9),则x=t²-9 不定积分可化为∫t/(t²-9)d(t²-9)dt=∫2t²/(t²-9)dt=∫[2+3/(t-3)-3/(t+3)]dt=2t+3ln(t-3)-3ln(t+3)

令x=3sint 原式=∫(-π/2→π/2)(1+3sint)*3cost*3costdt=∫(-π/2→π/2)9cos^2(t)dt+∫(-π/2→π/2)27sintcos^2(t)dt=9/2∫(-π/2→π/2)(cos(2t)+1)dt-27∫(-π/2→π/2)cos^2(t)d(cost)=9/4sin(2t)|(-π/2→π/2)+9/2t|(-π/2→π/2)-9cos^3(t)|(-π/2→π/2)=0+9/2π-0=4.5π

网站首页 | 网站地图
All rights reserved Powered by www.prss.net
copyright ©right 2010-2021。
内容来自网络,如有侵犯请联系客服。zhit325@qq.com