Join the Most Relevant JEE Main 2025 Test Series & get 99+ percentile! Join Now
Search any question & find its solution
Question: Answered & Verified by Expert
If m is any natural number, then the value of the integral x3m+x2m+xm2x2m +3xm+61/mdx is (where, C is an arbitrary constant)
MathematicsIndefinite IntegrationJEE Main
Options:
  • A 16m+12x3m+3x2m+6xm1/m+1 +C
  • B 16m2x3m+3x2m+6xm 1/m+1 +C
  • C 16m2x3m+3x2m+6xm1/m+C
  • D None of the above
Solution:
1698 Upvotes Verified Answer
The correct answer is: 16m+12x3m+3x2m+6xm1/m+1 +C

Put, I=x3m+x2m+xm2x3m+3x2m+6xm1/mxdx

=x3m-1+x2m-1+xm-12x3m +3x2m +6xm1/mdx

=16m2x3m+3x2m+6xm1/m6mx3m-1 +6mx2m-1 +6mxm-1dx

Now, let 2x3m+3x2m+6xm=t6mx3m-1 +6mx2m-1 +6mxm-1dx=dt 

I=16mt1/mdt =16mt1/m+11/m+1

=16m+12x3m+3x2m+6xm(1/m)+1+C

Looking for more such questions to practice?

Download the MARKS App - The ultimate prep app for IIT JEE & NEET with chapter-wise PYQs, revision notes, formula sheets, custom tests & much more.