谁能帮我解一个定积分的题目，谢谢

令x = sec t, x = 5/4时, t = arccos 4/5, x = 1时, t = 0
dx = tan t sec t dt
x² - 1 = tan² t
√(x² - 1) = tan t
代入原式得
∫√(x² - 1) dx = ∫tan t tan t sec t dt
=∫sin²t/cos³t dt
=∫sin²t cost/(cost)^4 dt
=∫sin²t/(1-sin²t)² dsint
令u = sin t
则原式
=∫u²/(1-u²)² du
=∫[u/(1-u²)]² du
=∫1/4 * [1/(1-u) - 1/(1+u)]² du
=1/4 ∫[1/(1-u)² + 1/(1+u)² - 1/(1-u) - 1/(1+u)] du
=1/4 [1/(1-u) - 1/(1+u) + ln(1-u) - ln(1+u)]
=1/4 [2sint/cos²t + 2ln(1-sint) - 2lncost]
=1/4 [2√(1 - 1/x²) / 1/x² + 2ln(1 - √(1 - 1/x²)) - 2ln(1/x)] + C
=x/2 √(x²-1) + 1/2 ln[x - √(x²-1)] + C