∫1x2x4+134dx=
-x4+134x+C
-1-x4142x+C
-1+x414x+C
-1+x414x2+C
-1+x412x+C
Explanation for Correct answer:
Finding the value for the given integral:
∫1x2x4+134dx=∫1x2x4341+1x434dx=∫1x2x31+1x434dx=∫x-51+x-434dx
Solving this by using the substitution method
t4=1+x-44t3dt=-4x-5dx∵ddxxn=nxn-1t3dt=-x-5dx
∫x-51+x-434dx=-∫t3dtt434=-∫dt=-t+C=-1+x-414+C=-1+1x414+C=-x4+1x414+C=-x4+114x414+C=-x4+114x+C
Hence, option (C) is the correct answer.
Write = or ≠ in the place holder.18□34
Consider two events A and B such that P(A)=14, P(BA)=12, P(AB)=14. For each of the following statements, which is true.
I.P(A'B')=34
II. The events A and B are mutually exclusive
III.P(AB)+P(AB')=1