∫2dxex+e-x2=
-e-xe-x+ex+C
1e-x+ex+C
1ex+12+C
1e-x+ex2+C
Explanation for Correct answer:
Evaluating the given integral:
∫2ex+e-x2=∫2ex+1ex2dx=∫2exex+1ex2dx=∫2exe2x+12dx
Solving this by using the substitution method
Let t=e2x+1⇒dt=2e2xdx
∫2ex+e-x2=∫dtt2=-1t+C=-1e2x+1+C=-e-xe-x+ex+C
Hence, option (C) is the correct answer.
The general solution of a differential equation of the type dxdy+P1x=Q1 is
(a) ye∫P1 dy=∫(Q1e∫P1 dy)dy+C
(b) ye∫P1 dx=∫(Q1e∫P1 dx)dx+C
(c) xe∫P1 dy=∫(Q1e∫P1 dy)dy+C
(d) xe∫P1 dx=∫(Q1e∫P1 dx)dx+C