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Solving: ∫−x2+3x+7(x+2)2(x2+x+1)2dx\int\frac{-x^2+3x+7}{\left(x+2\right)^2\left(x^2+x+1\right)^2}dx
Solve instead: ∫−x2+3x+7(x+2)2(x2+x+1)2dt\int\frac{-x^2+3x+7}{\left(x+2\right)^2\left(x^2+x+1\right)^2}dt
f(x)=(-x^2+3x+7)/((x+2)^2(x^2+x+1)^2)−6−5−4−3−2−10123456−3-2.5−2-1.5−1-0.500.511.522.53xy

Exercise

∫−x2+3x+7(x+2)2(x2+x+1)2dt\int\frac{-x^2+3x+7}{\left(x+2\right)^2\left(x^2+x+1\right)^2}dt

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Final answer to the exercise

13(x+2)+4ln⁡∣x+1∣−4ln⁡∣x∣+−83(x+1)+4−3x+19ln⁡∣x+2∣+3arctan⁡(1+2x3)3−19ln⁡∣(x+12)2+34∣+C2\frac{1}{3\left(x+2\right)}+4\ln\left|x+1\right|-4\ln\left|x\right|+\frac{-8}{3\left(x+1\right)}+\frac{4}{-3x}+\frac{1}{9}\ln\left|x+2\right|+\frac{\sqrt{3}\arctan\left(\frac{1+2x}{\sqrt{3}}\right)}{3}-\frac{1}{9}\ln\left|\sqrt{\left(x+\frac{1}{2}\right)^2+\frac{3}{4}}\right|+C_2

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∫−x2+3x+7(x+2)2(x2+x+1)2 dt
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