Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It only takes a minute to sign up.
Sign up to join this community
I am having doubt in the calculation of the integral $$ \int\frac{1}{1+x} \ dx,$$
the solution of which is
$$
\log(1+x)+C.$$
I have solved this integration in a different way. First I converted the above integral to
$$
\;\int\frac{1}{1+(\sqrt{x})^2} \ dx.$$
Then I used the formula as
$$
{\int\frac{1}{1+x^2} dx}=\tan^{-1}x+C
$$
so by using this formula I got as an answer $\tan^{-1}(\sqrt{x})+C$ which is different from the solution.
Am I correct about this?
No. You forgot to change $\rm{d}x$ into $\rm{d}(\sqrt x)$.
The substitution $x=t^2$ (which, by the way, can be done only for $x\ge0$) brings the integral in the form $$ \int\frac{1}{1+t^2}\cdot 2t\,dt=\log(1+t^2)+c=\log(1+x)+c $$
Recall that integration by substitution is an application of the chain rule for derivatives. The $dx$ is a reminder for you to apply it (more than that, actually).
Otherwise you'd get, similarly to your manipulation, $$ \int x\,dx=\int (\sqrt{x})^2\,dx=\frac{(\sqrt{x})^3}{3}+c $$ which is clearly absurd. Or $$ \int\frac{1}{\sqrt{1-x}}\,dx=\int\frac{1}{\sqrt{1-(\sqrt{x})^2}}\,dx=\arcsin\sqrt{x}+c $$ whereas the correct antiderivative is $-2\sqrt{1-x}+c$.
