integral of dx/(x^2+1)x^0.5

proceed this way

Dear Shivan,

 put x = t²

dx = 2tdt

so intergral is -   2dt/(t⁴+1)

 =  2 [ (t²+1) - (t²-1) ] /(t⁴+1)  dt

solve intergral (t²+1)/(t⁴+1) = (1+ 1/t²)/[(t-1/t)² + 2]

now pur t-1/t = y or (1+ 1/t²)dt = dy

put above value in integral and solve

similarly, solve intergral : (t²-1)/(t⁴+1)

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Askiitians Expert

Anil Pannikar

IIT Bombay

    Hello,

            Iam Surya Anuraag.Here is your answer,

            dx/(x^2+1)x^0.5 ,

            put x^0.5 = t , dx/2(x)^0.5=dt, => dx/x^0.5=2dt

             now  it wll be in the form,

              2dt/(t^4+1)=(t^2+1)dt/(t^4+1) - (t^2-1)dt/(t^4+1)

              now you can simplify quite easily by takung x^2 common from both nm.,&dm.,   from two fractions and by putting the integrals of the nm`s.,as a function

                                                                         Thankyou.

is it 4/3 log x

Thanks and Regards 
Sher Mohammad
Askiitians Faculty,
B.tech, IIT Delhi