How to transform $\int_0^\infty x(1+x^2)^{-2}dx$ to $0,1$ as limits of the integral?

Question

Consider $\displaystyle\int_0^\infty x(1+x^2)^{-2}dx$

If I consider $y=\frac{1}{1+x^2}$, then $dy=\frac{-dx}{(1+x^2)^2}$

and $x=\pm\sqrt{\frac{1}{y}-1}$

and then $\displaystyle\int_1^0 \mp\sqrt{\frac{1}{y}-1}dy$

however this is equal to $\pi/2$ and the answer of the book is 1/2 as the exact answer and 1 when is transformed on the interval $(0,1)$

Why?

What am I doing wrong?

Answer 1

You changed the variables wrong. The correct expression for $dy$ comes from: $$y=\frac1{1+x^2}\\ dy=\frac{-2x\,dx}{(1+x^2)^2}$$ Then when you substitute this into the integral, get $$\int_{1}^0-\frac12\,dy=\frac12$$

Answer 2

I think you're just missing an $x$ when you transform your integral to y. Should be $dy=\frac{-2x}{(1+x^2)^2}dx$. (You forgot the "inside" derivative).

Alternatively, try the substitution $y=x^2+1$ to make derivatives easier.