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

Exercise

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

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Final answer to the exercise

74(x+2)+316ln⁡∣x∣+57arctan⁡(3+2x7)16+916ln⁡∣(x+32)2+74∣−34ln⁡∣x+2∣+C2\frac{7}{4\left(x+2\right)}+\frac{3}{16}\ln\left|x\right|+\frac{5\sqrt{7}\arctan\left(\frac{3+2x}{\sqrt{7}}\right)}{16}+\frac{9}{16}\ln\left|\sqrt{\left(x+\frac{3}{2}\right)^2+\frac{7}{4}}\right|-\frac{3}{4}\ln\left|x+2\right|+C_2

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∫ x2+3(x+2)2x(x2+3x+4) dx
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