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f(x)=(1x+1)/(x^2(x-1)^3(x^2+4))−6−5−4−3−2−10123456−3-2.5−2-1.5−1-0.500.511.522.53xy

Exercise

∫1x+1x2(x−1)3(x2+4)dx\int\frac{1x+1}{x^2\left(x-1\right)^3\left(x^2+4\right)}dx

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Final answer to the exercise

14x+−15(x−1)2−71000arctan⁡(x2)−3125ln⁡∣x2+4∣−ln⁡∣x∣+1.024ln⁡∣x−1∣+1925(x−1)+C1\frac{1}{4x}+\frac{-1}{5\left(x-1\right)^{2}}-\frac{7}{1000}\arctan\left(\frac{x}{2}\right)-\frac{3}{125}\ln\left|\sqrt{x^2+4}\right|-\ln\left|x\right|+1.024\ln\left|x-1\right|+\frac{19}{25\left(x-1\right)}+C_1

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∫1x+1x2(x−1)3(x2+4) dx
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