integrate 1/(e^x-1)^2
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Answer:
1/2(e^x-1)ln(e^x-1)^2+c
Step-by-step explanation:
Put (e^x-1)=t
=2(e^x-1)dx=dt
=dx=dt/2(e^x-1)
=In( 1/t dt/2(e^x-1)
=1/2(e^x-1)ln(e^x-1)^2+c
Pls tell if my ans is correct or not??
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