A) \[-{{[f(x)]}^{-1}}+c\]
B) \[\log [f(x)]+c\]
C) \[{{e}^{f(x)}}+c\]
D) None of these
Correct Answer: A
Solution :
Put \[f(x)=t\Rightarrow {f}'(x)\,dx=dt,\] then \[\int_{{}}^{{}}{\frac{{f}'(x)}{|f(x){{|}^{2}}}\,dx}=\int_{{}}^{{}}{\frac{1}{{{t}^{2}}}\,dt}=-\frac{1}{t}+c=-\frac{1}{f(x)}+c.\]
You need to login to perform this action.
You will be redirected in
3 sec
