∫x²/√1-x²
2个回答
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令x=sint，则t=arcsinx
∫[x²/√(1-x²)]dx
=∫sin²t/√(1-sin²t)]d(sint)
=∫(sin²t·cost/cost)dt
=∫sin²tdt
=½∫(1-cos2t)dt
=½t-¼sin2t +C
=½arcsinx-½x√(1-x²)+C
=½[arcsinx-x√(1-x²)] +C
∫[x²/√(1-x²)]dx
=∫sin²t/√(1-sin²t)]d(sint)
=∫(sin²t·cost/cost)dt
=∫sin²tdt
=½∫(1-cos2t)dt
=½t-¼sin2t +C
=½arcsinx-½x√(1-x²)+C
=½[arcsinx-x√(1-x²)] +C
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(3)
∫(x²+1)/[(x+1)²(x-1)]dx
=∫[x/(x²-1) -1/(x+1)²]dx
=∫[x/(x²-1)]dx -∫[1/(x+1)²]dx
=½∫[2x/(x²-1)]dx - ∫[1/(x+1)²]d(x+1)
=½ln|x²-1| +(x+1)⁻¹ +C
∫(x²+1)/[(x+1)²(x-1)]dx
=∫[x/(x²-1) -1/(x+1)²]dx
=∫[x/(x²-1)]dx -∫[1/(x+1)²]dx
=½∫[2x/(x²-1)]dx - ∫[1/(x+1)²]d(x+1)
=½ln|x²-1| +(x+1)⁻¹ +C
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