∫logx+1+x21+x2dx=
logx+1+x22+c
xlogx+1+x2+c
12logx+1+x2+c
12logx+1+x22+c
x2logx+1+x2+c
Explanation for the correct option:
Finding the value of given integration:
Consider the given equation as,
I=∫logx+1+x21+x2dx
Let us assume that,
logx+1+x2=t⇒dx1+x2=dt [differentiating]
Then,
I=∫tdtI=t22+c
Substitute the value of the t in the above Equation
I=logx+1+x222+cI=12logx+1+x22+c
Hence, the correct answer is Option (D).
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