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```        Sir,
how to integrate
dx/[k-m{(2nx)^1/2}]
from 0 to x
k,m&n are constants```
7 years ago

147 Points
```										Dear aman
o∫xdx/[k-m√2nx]
we can also wrote

I = 1/m√2n o∫xdx/[(k/m√2n)-√x]
=  1/m√2n o∫xdx/[a-√x]            where a =(k/m√2n)
I= 1/m√2n o∫x[a+√x]dx/[a-x]
I= 1/m√2n o∫x[a]dx/[a-x]  +  1/m√2n o∫x[√x]dx/[a-x]
first part u can easly calculate
for second part
I2 = o∫xxdx/√x[a-x]
put √x = t
dx = 2tdt
I2 = o∫√x2t2dt/[a-t2]
this is also a standard form
Please feel free to post as many doubts on our discussion forum as you can.If you find any question Difficult to understand - post it here and we will get youthe answer and detailed  solution very  quickly. We are all IITians and here to help you in your IIT JEE preparation.All the best. Regards,Askiitians ExpertsBadiuddin

```
7 years ago
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