Question

不定积分2道(过程)

1，∫dx/x^2-3x-4
,2， ∫ tan^5 x dx

Answer 1

LZ写的方式有歧义，在下说下自己的看法和对应的解法
1，∫ dx/(x^2-3x-4)
原式=∫ dx/[(x-4)*(x+1)]
=∫ [dx/(x-4)-dx/(x+1)]/5
=[∫ dx/(x-4)-∫ dx/(x+1)]/5
=[ln(x-4)-ln(x+1)]/5
=ln[(x-4)/(x+1)]/5
我想加上括号就不会产生歧义了吧
2，∫ (tanx)^5＊dx
原式=∫ (sin x/cos x)^5＊dx
=∫ [(sin x)^5/(cos x)^5]dx
上下同时乘以cos x
=∫ [(sin x)^5*(cos x)*dx]/(cos x)^6
写到这步你应该猜出来我要第一次换元了
=∫ [(sin x)^5*d(sin x)]/[1-(sin x)^2]^3
令t=sin x
=∫ (t^5*dt)/(1-t^2)^3
=∫ [t^4*d(t^2/2)]/(1-t^2)^3
令u=t^2=(sin x)^2
=∫ (u^2*du)/[2*(1-u)^3]
令v=1-u=(cos x)^2
=∫ -[(1-v)^2*dv]/[2*v^3]
展开
=∫ [-dv/(2*v)+dv/(v^2)-dv/(2*v^3)]
各项分别积分
= -(ln v)/2-1/v-1/(4*v^2)
代入v=(cos x)^2
= -ln(cos x)-(sec x)^2-(sec x)^4/4
自己验算一下吧，两年没做过数学题了

Answer 2

1。1/(x² - 3x - 4) = 1/[(x + 1)(x - 4)] = A/(x + 1) + B/(x - 4)
1 = A(x - 4) + B(x + 1)
代入x = -1，1 = A(-5) => A = -1/5
代入x = 4，1 = B(5) => B = 1/5
∴1/(x² - 3x - 4) = -1/[5(x + 1)] + 1/[5(x - 4)]
∫ dx/(x² - 3x - 4) = (-1/5)∫ dx/(x + 1) + (1/5)∫ dx/(x - 4)
= (-1/5)ln|x + 1| + (1/5)ln|x - 4| + C
= (1/5)ln|(x - 4)/(x + 1)| + C

2。∫ (tanx)^5 dx
= ∫ tan³x (sec²x - 1) dx，恒等式：1 + tan²x = sec²x
= ∫ tan³x sec²x - ∫ tan³x dx
= ∫ tan³x d(tanx) - ∫ tanx (sec²x - 1) dx
= (1/4)tan⁴x - ∫ tanx sec²x + ∫ tanx dx
= (1/4)tan⁴x - ∫ tanx d(tanx) + ∫ sinx/cosx dx
= (1/4)tan⁴x - (1/2)tan²x - ln|cosx| + C