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/

1个回答

  • (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