∫[(9+X^2)^(1⼀2)]dx

∫ √(9 + x²) dx = ∫ √(3² + x²) dx，于是选用x = 3tanθ，dx = 3sec²θ dθ
= ∫ √(9 + 9tan²θ) * 3sec²θ dθ
= 9∫ sec³θ dθ
= (9/2)[secθtanθ + ln(secθ + tanθ)] + C
= (9/2)(x/3)[√(9 + x²)/3] + (9/2)ln[x/3 + √(9 + x²)/3] + C
= (1/2)x√(9 + x²) + (9/2)ln[x + √(9 + x²)] + C

下面是∫ sec³x dx的做法：
∫ sec³x dx = ∫ secx * sec²x dx = ∫ secx d(tanx)
= secxtanx - ∫ tanx d(secx)
= secxtanx - ∫ tanx * secxtanx dx
= secxtanx - ∫ secx * (sec²x - 1) dx
= secxtanx - ∫ sec³x dx + ∫ secx dx
2∫ sec³x dx = secxtanx + ∫ secx * (secx + tanx)/(secx + tanx) dx
∫ sec³x dx = (1/2)secxtanx + (1/2)∫ d(secx + tanx)/(secx + tanx)
∫ sec³x dx = (1/2)secxtanx + (1/2)ln(secx + tanx)

求不定积分∫[√(9+X²)]dx
原式=3∫√[1+(X/3)²)dx 【令x/3=tanu，则x=3tanu，dx=3sec²udu，代入原式得】
=9∫sec³udu=9∫secud(tanu)=9[secutanu-∫secu(sec²u-1)du]
=9[secutanu-∫sec³udu+∫secudu]=9[secutanu-∫sec³udu+ln(secu+tanu)]+2C₁
移项得18∫sec³udu=9[secutanu+ln(secu+tanu)]+2C₁
两边同除以2得 ∫[√(9+X²)]dx=9∫sec³udu=(9/2)[secutanu+ln(secu+tanu)]+C₁
=(9/2){(x/9)√(9+X²)+ln[(1/3)√(9+X²)+x/3]}+C₁
=(x/2)√(9+X²)+(9/2)ln[√(9+X²)+x]+C，其中C=C₁-(9/2)ln3.

令x=3tant
dx=3sec^2tdt
代入原式得
原式
=∫3sect*3sec^2tdt
=9∫sec^3tdt
=∫3sect*d(3tant)
=9secttant-9∫tantdsect
=9secttant-9∫tant*secttantdt
=9secttant-9∫sect(sec^2t-1)dt
=9secttant-9∫sec^3tdt+9∫sectdt
因此
18∫sec^3tdt

=9secttant+9∫sectdt
=9secttant+9∫1/costdt
=9secttant+9∫cost/cos^2tdt
=9secttant+9∫1/(1-sin^2t)dsint
=9secttant+9/2∫[1/(1-sint)+1/(1+sint)]dsint
=9secttant+9/2ln[(1+sint)/(1-sint)]+C
原式
=9∫sec^3tdt
=9/2secttant+9/4ln[(1+sint)/(1-sint)]+C
自己反代吧