∫x²+1/(x+1)²(x-1)dx
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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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