Join now for JEE/NEET and also prepare for Boards Learn Science & Maths Concepts for JEE, NEET, CBSE @ Rs. 99! Register Now
Hey there! We receieved your request
Stay Tuned as we are going to contact you within 1 Hour
One of our academic counsellors will contact you within 1 working day.
Click to Chat
1800-5470-145
+91 7353221155
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 )
Sit and relax as our customer representative will contact you within 1 business day
OTP to be sent to Change
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
Get your questions answered by the expert for free
You will get reply from our expert in sometime.
We will notify you when Our expert answers your question. To View your Question
Win Gift vouchers upto Rs 500/-
Register Yourself for a FREE Demo Class by Top IITians & Medical Experts Today !