∫ (x-2)/ ((x+1)^2+4) dx
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∫ (x-2)/ ((x+1)^2+4) dx
![∫ (x-2)/ ((x+1)^2+4) dx](/uploads/image/z/4693072-40-2.jpg?t=%E2%88%AB+%28x-2%29%2F+%28%28x%2B1%29%5E2%2B4%29+dx)
令x+1=t,则x=t-1,dx=dt
原式=∫ (t-3)/(t^2+4)dx =∫ t/(t^2+4)dt-3∫ 1/(t^2+4)d
=(1/2)∫ 1/(t^2+4)d(t^+4)-3arctan(t/2)+C
=(1/2)ln(t^2+4)-3arctan(t/2)+C
=(1/2)ln[(x+1)^2+4]-3arctan[(x+1)/2]+C
原式=∫ (t-3)/(t^2+4)dx =∫ t/(t^2+4)dt-3∫ 1/(t^2+4)d
=(1/2)∫ 1/(t^2+4)d(t^+4)-3arctan(t/2)+C
=(1/2)ln(t^2+4)-3arctan(t/2)+C
=(1/2)ln[(x+1)^2+4]-3arctan[(x+1)/2]+C