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∫dx/(x²+2)(x+1)]=(1/3)∫[-(x-1)/(x²+2)+1/(x+1)]dx
=(1/3)[∫-xdx/(x²+2)+∫dx/(x²+2)+∫dx/(x+1)]
=(1/3)[-(1/2)∫d(x²+2)/(x²+2)+∫dx/(x²+2)+∫dx/(x+1)]
=(1/3)[(-1/2)ln(x²+2)+(1/√2)arctan(x/√2)+ln∣x+1∣]+C
=-(1/6)ln(x²+2)+[(√2)/6]arctan(x/√2)+(1/3)ln∣x+1∣]+C;
=(1/3)[∫-xdx/(x²+2)+∫dx/(x²+2)+∫dx/(x+1)]
=(1/3)[-(1/2)∫d(x²+2)/(x²+2)+∫dx/(x²+2)+∫dx/(x+1)]
=(1/3)[(-1/2)ln(x²+2)+(1/√2)arctan(x/√2)+ln∣x+1∣]+C
=-(1/6)ln(x²+2)+[(√2)/6]arctan(x/√2)+(1/3)ln∣x+1∣]+C;
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