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Grade: 11
        
Ans : 2log(e^x +1) -x 
How do I solve this problem?????
3 months ago

Answers : (1)

Aditya Gupta
1690 Points
							
write (e^x – 1)/(e^x+1)= 2e^x/(e^x+1) – 1
now integrating
∫ 2e^x/(e^x+1) dx – ∫1 dx
= ∫ 2e^x/(e^x+1) dx – x
to find ∫ 2e^x/(e^x+1) dx, put y= e^x+1 so that dy= e^x dx
so ∫ 2e^x/(e^x+1) dx= ∫ 2dy/y= 2*∫dy/y= 2 log y= 2log(e^x +1) 
so, the integral is equal to 
2log(e^x +1) – x
kindly approve :)
3 months ago
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