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
-11log(1+x)1+x2dx=01log(1+x)1+x2dx+01fxdx then fx=
MathematicsDefinite IntegrationTS EAMCETTS EAMCET 2021 (04 Aug Shift 1)
Options:
  • A log(1+x)1+x2
  • B -log(1+x)1+x2
  • C log(1-x)1+x2
  • D 0
Solution:
2832 Upvotes Verified Answer
The correct answer is: log(1-x)1+x2

Given -11log(1+x)1+x2dx=-10log(1+x)1+x2dx +01log(1+x)1+x2dx

 abfxdx=abfxdx+bcfxdx

Now, 

Let I1=-10log1+x1+x2dx

Put x=-tdx=-dt

when x=-1, t=1 & when x=0, t=0

 I1=10log1-t1+t2-dtI1=01log1-t1+t2dt     abfxdx=-bafxdx

 I1=01log1-x1+x2dx  abfxdx=abftdt

 -11log1+x1+x2dx=01log1+x1+x2dx+01fxdx

=01log1+x1+x2dx+01log1-x1+x2dx

 fx=log1-x1+x2

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.