A curve with gradient function (1-x)/(1+x^2) goes through the origin. Show that the curve cuts the X-axis when arctan(x) = 0.5ln(1+x^2)

What exactly am I meant to do for these types of questions?

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- Nov 15th 2009, 08:11 PMSunyataInverse Trigonometry.
A curve with gradient function (1-x)/(1+x^2) goes through the origin. Show that the curve cuts the X-axis when arctan(x) = 0.5ln(1+x^2)

What exactly am I meant to do for these types of questions? - Nov 15th 2009, 08:20 PMredsoxfan325
To find the original function you need to integrate

$\displaystyle \int\frac{1-x}{1+x^2}\,dx=\int\frac{1}{1+x^2}\,dx-\int\frac{x}{1+x^2}\,dx$

The first integral is $\displaystyle \arctan(x)$ (you should find this on any table of integrals) and the second integral can be solved using a simple u-sub ($\displaystyle u=1+x^2$), so you end up with $\displaystyle f(x)=\arctan(x)-\frac{1}{2}\ln(1+x^2)$. Then you set $\displaystyle f(x)=0$ to get the answer.

It seems a little strange that the question would ignore the presence of the integration constant though.