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`        integral of dx/(x^2+1)x^0.5`
7 years ago

souvik das
33 Points
```										proceed this way

```
7 years ago
85 Points
```										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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IIT Bombay

```
7 years ago
Surya Anuraag Duvvuri
39 Points
```										    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.
```
7 years ago
44 Points
```										is it 4/3 log x
```
7 years ago
IIT Delhi
174 Points
```										Thanks and Regards Sher MohammadAskiitians Faculty,B.tech, IIT Delhi
```
3 years ago
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