不定积分2道(过程)1,∫dx/x^2-3x-4,2, ∫ tan^5 x dx

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

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...