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

souvik das
33 Points
13 years ago

proceed this way

85 Points
13 years ago

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)

All the best.

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Surya Anuraag Duvvuri
39 Points
13 years ago

Hello,

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.

44 Points
13 years ago

is it 4/3 log x

10 years ago

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