Question

∫[0,1]√[x(4+5x)]dx定积分感谢

Answer 1

先推导不定积分公式∫√（x^2-a^2)dx,
设x=asect,dx=asect*tantdt,
sect=x/a,tant=(1/a)√（x^2-a^2)
原式=∫atant*asect*tantdt
=a^2∫（tant)^2sectdt
=a^2∫[（sect)^3-sect]dt,
∫（sect)^3=∫sectd(tant)
=sect*tant-∫tant*d(sect)
=sect*tant-∫sect*(tant)^2dt
=sect*tant-∫(sect)^3dt+∫sectdt,
∫（sect)^3dt=(1/2)sect*tant+(1/2)∫sectdt,
原式=(1/2)sect*tant-(1/2)∫sectdt,
=(1/2)sect*tant-(1/2)ln|sect+tant|+C1
=a^2*(1/2)[(x/a)(1/a)√（x^2-a^2)-a^2(1/2)ln|(x/a)+(1/a)√（x^2-a^2)|+C1
=(x/2)√（x^2-a^2)-(a^2/2)ln|x+√（x^2-a^2)+C,
∫[0,1]√[x(4+5x)]dx
=√5*∫[0,1]√[(x+2/5)^2-(2/5)^2]d(x+2/5)
=√5[0,1] {(x+2/5)/2√[(x+2/5)^2-(2/5)^2]-[(2/5)^2/2]ln[x+2/5+√[(x+2/5)^2-(2/5)^2]
=21/10-(2√5/25)ln[(7+3√5）/5]+2√5/25ln(7/5).
=21/10+（2√5/25）ln[(49-21√5）/4]。