# `dy/dx` = `(10x)/(1+x^2)^2`   Find `intx/(1+x^2)^2 dx`

Borys Shumyatskiy | Certified Educator

Make the substitution y = 1 + x^2, then dy = 2x*dx.

So the integral is equal to

∫(1/2)*(1/y^2)dy = (1/2)*(-1)/y = -(1/2)/(1+x^2) + C.

(for the y mentioned at the first line, y(x) = -5/(1+x^2) + C.

kspcr111 | Student

CONT...

Now let us find the `int x/(1+x^2)^2 dx`

let `u= (1+x^2): du = 2xdx , dx=1/2x *du `

`= int (x/u^2)*(1/2x) dx `

`=int 1/(2u^2) du`

`= 1/2 int u^-2 du`

`= 1/2 (u^-1/-1) `

`= -1/2(1+x^2) `

in the first problem we got ` y= -5/(1+x^2) `

`=> y/5 = -1/(1+x^2) `

`=> y/10 =-1/(2*(1+x^2)) `

so , `int x/(1+x^2)^2 dx = y/10`

kspcr111 | Student

The first one is simple and is as follows

`dy/dx =(10*x)/(1+x^2)^2`

Integrating on both sides we get

`=> y= int((10*x)/(1+x^2)^2)dx`

= `10*int(((*x)/(1+x^2)^2)dx)`

Let u=(1+x^2)   =>  du =2x *dx   , dx = `(1/(2x)) * du`

` = 10 int (x/u^2)*(1/2x)*du`

`= 10 int(1/(2*u^2))* du `

` = 10/2(int (1/u^2)) * du`

` = 5 *((u^ (-1))/-1)`

` = -5/u`

`=-5/(1+x^2) `

` => y= (-5)/(1+x^2)`

` `

stacy143 | Student

`dy/dx =10*x/(1+x^2)^2 `

=> `(1/10)*(dy/dx)= x/(1+x^2)^2 `

Now integrating on both sides we get

` int (1/10)*(dy)= int (x/(1+x^2)^2) dx`

` => int (x/(1+x^2)^2) dx = y/10`

Hope this helps you