Question

int 4x+1/x+1 evaluation​

Answer 1

Answer:

sorry bro of the correct answer

Answer 2

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