求一个不定积分的题目，谢谢
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设 tant = x+1，则 (sect)^2*dt = dx
∫(x+2)*dx/橘御拦拍(x^2 +2x +2)
=∫(x+2)*dx/[(x^2 +2x +1) +1]
=∫(x+2)*dx/[(x+1)^2 +1]
=∫(tant +1)*(sect)^2*dt/[(tant)^2 + 1]
=∫(tant + 1)*(sect)^2 *dt/(sect)^2
=∫tant *dt + ∫dt
=∫圆衡岩sint*dt/cost + t
=-∫d(cost)/cost + t
=-ln|cost| + t + C
=-ln|1/sect| + arctan(x+1) +C
=ln|sect| + arctan(x+1) + C
=1/2*ln|(sect)^2| + arctan(x+1) + C
=1/2*ln|1 + (1+x)^2| + arctan(x+1) +C
=1/2*ln|x^2 +x +2| + arctan(x+1) + C
∫(x+2)*dx/橘御拦拍(x^2 +2x +2)
=∫(x+2)*dx/[(x^2 +2x +1) +1]
=∫(x+2)*dx/[(x+1)^2 +1]
=∫(tant +1)*(sect)^2*dt/[(tant)^2 + 1]
=∫(tant + 1)*(sect)^2 *dt/(sect)^2
=∫tant *dt + ∫dt
=∫圆衡岩sint*dt/cost + t
=-∫d(cost)/cost + t
=-ln|cost| + t + C
=-ln|1/sect| + arctan(x+1) +C
=ln|sect| + arctan(x+1) + C
=1/2*ln|(sect)^2| + arctan(x+1) + C
=1/2*ln|1 + (1+x)^2| + arctan(x+1) +C
=1/2*ln|x^2 +x +2| + arctan(x+1) + C
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