Question

求下列不定积分

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。

~求详解（解题过程）~谢谢@

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