int 4x+1/x+1 evaluation
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Answer:
sorry bro of the correct answer
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Explanation:
We have , I=∫4x+1x2+3x+2dx
⇒I=∫2(2x+3)−5x2+3x+2dx [Using4x+1=λ(2x+3)+μ]
⇒=2∫2x+3x2+3x+2dx−5∫1x2+3x+2dx
⇒I=2∫1x2+3x+2d(x2+3x+2)−5∫1(x+32)2−(12)2dx
⇒I=2log∣∣x2+3x+2∣∣−5log∣∣∣x+1x+2∣∣∣+C
⇒I=2log|x+1|+2log|x+2|−5log|x+1|+5log|x+2|+C
⇒I=−3log|x+1|+7log|x+2|+C
∴a=−3andb=7⇒a+b=4
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