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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?

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    $\begingroup$ Differentiating $\tan^{-1}\sqrt x$ gives $\frac1{2\sqrt x(1+x)}$. $\endgroup$
    – Angina Seng
    Dec 26 '18 at 18:48
  • $\begingroup$ The correct answer is $\ln(|1+x|)+C$. $\endgroup$
    – hamam_Abdallah
    Dec 26 '18 at 20:02
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No. You forgot to change $\rm{d}x$ into $\rm{d}(\sqrt x)$.

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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 $-2\sqrt{1-x}+c$.

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