Falling Behind in Studies? Join Our Performance Improvement Batch. Enroll For Free
One of our academic counsellors will contact you within 1 working day.
Click to Chat
1800-1023-196
+91-120-4616500
CART 0
Use Coupon: CART20 and get 20% off on all online Study Material
Welcome User
OR
LOGIN
Complete Your Registration (Step 2 of 2 )
dx/(x-1)(x^2+4(under the root))
Integral dx/(x-1).sqrt of (x^2+4) Put x-1=1/t On diff w.r. to x, we get dx=-1/t^2dt =integral -1/t^2.dt/(1/t).sqrt of ((1/t +1)^2+4) = -integral dt/root of (5t^2+2t+1) =-1/root5 integral dt/sqrt of ((t+1/5)^2+(4/5)^2 =-1/root5log mod {(t+1/5) +sqrt of (t^2+2t/5+1/5)} +C =-1/root5 log mod [ {1/(x-1)+1/5}+sqrt of {1/(x-1)^2 +2/5.(x-1) +1/5}]+C =-1/root5 log mod [{1/(x-1)+1/5} + sqrt of{ (x^2+4)/5.(x-1)^2}] +CThanks & Regards Rinkoo Gupta AskIITians Faculty
Integral dx/(x-1).sqrt of (x^2+4)
Put x-1=1/t
On diff w.r. to x, we get dx=-1/t^2dt
=integral -1/t^2.dt/(1/t).sqrt of ((1/t +1)^2+4)
= -integral dt/root of (5t^2+2t+1)
=-1/root5 integral dt/sqrt of ((t+1/5)^2+(4/5)^2
=-1/root5log mod {(t+1/5) +sqrt of (t^2+2t/5+1/5)} +C
=-1/root5 log mod [ {1/(x-1)+1/5}+sqrt of {1/(x-1)^2 +2/5.(x-1) +1/5}]+C
=-1/root5 log mod [{1/(x-1)+1/5} + sqrt of{ (x^2+4)/5.(x-1)^2}] +C
Thanks & Regards
Rinkoo Gupta
AskIITians Faculty
Post Question
Dear , Preparing for entrance exams? Register yourself for the free demo class from askiitians.
points won -