Question

不定积分∫(1/x√x²-1)dx -1 -2

求下列不定积分：
(1)∫[1/(√1-x^2)-2/(1+x^2)-5/x^2]dx
计算下列不定积分：
(2)∫√x[5/(x√x)+(1/x^2)]dx
(3)∫3/[X^2(1+x^2)]dx
(4)∫2^x[(e^x)-1]dx
(5)∫[(e^2x)-4]/[e^x+2] dx
(6)∫cos^2(t/2) dt
(7)∫cos2x/(cosx-sinx) dx
注：为了表述关系,原题中"()"=在此写成"[]",原题中无在此表述的"()"符号.

Answer 1

(1) ∫[1/(√1-x²)-2/(1+x²)-5/x²]dx = arcsinx - 2arctanx + 5/x + C
(2) ∫√x[5/(x√x)+(1/x²)]dx = ∫[5/x+ 1/x^(3/2)]dx
= 5ln|x| - 2/√x + C
(3) ∫3/[x²(1+x²)]dx = 3∫[1/x² - 1/(1+x²)]dx = -3(1/x + arctanx) + C
(4) ∫2^x[(e^x)-1]dx = ∫[(2e)^x - 2^x]dx = (2e)^x/ln(2e) - 2^x/ln2 + C
(5) ∫[(e^2x)-4]/[e^x+2]dx = ∫[(e^x)²-2²]/[e^x+2]dx
= ∫(e^x - 2)dx = e^x - 2x + C
(6) ∫cos²(t/2)dt = ½∫(1+cost)dt = ½t + ½sint + C = ½(1 + sint) + C
(7) ∫cos2x/(cosx-sinx)dx = ∫(cos²x-sin²x)/(cosx-sinx)dx
= ∫(cosx+sinx)dx = sinx - cosx + C