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

Exercise

∫(x3+x2+3(x2+4)2(x2+x+1))dx\int\left(\frac{x^3+x^2+3}{\left(x^2+4\right)^2\left(x^2+x+1\right)}\right)dx

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Final answer to the exercise

−x8(x2+4)−116arctan⁡(x2)+1−2(x2+4)+73arctan⁡(1+2x3)39+113ln⁡∣(x+12)2+34∣−326arctan⁡(x2)+113ln⁡∣2x2+4∣+C2\frac{-x}{8\left(x^2+4\right)}-\frac{1}{16}\arctan\left(\frac{x}{2}\right)+\frac{1}{-2\left(x^2+4\right)}+\frac{7\sqrt{3}\arctan\left(\frac{1+2x}{\sqrt{3}}\right)}{39}+\frac{1}{13}\ln\left|\sqrt{\left(x+\frac{1}{2}\right)^2+\frac{3}{4}}\right|-\frac{3}{26}\arctan\left(\frac{x}{2}\right)+\frac{1}{13}\ln\left|\frac{2}{\sqrt{x^2+4}}\right|+C_2

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∫(x3+x2+3(x2+4)2(x2+x+1) )dx
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