How to evaluate
$$\int \frac{1}{\sqrt{1+\ln x}}\mathrm{d}x$$
I am trying to evaluate this integral by substituting $\ln x =u$. So, $dx$ $=$ $e^{u}du$. So, the integral will become
$$\int \frac{e^{u}}{\sqrt{1+u}}\mathrm{d}u$$
Now I am thinking of substituting $u = \tan(\theta)^2$. Now $\mathrm{d}u = 2\tan(\theta)\sec(\theta)^{2}\mathrm{d}\theta$ . So, finally the integral will become $$2\int e^{\tan^{2}\theta}\tan(\theta)\sec(\theta)\mathrm{d}\theta$$ Now if we again substitute $\sec(\theta)=z$ then $\sec(\theta)\tan(\theta)\mathrm{d}\theta = \mathrm{d}z$ . So, the next step of the expression will become $$\frac{2}{e}\int e^{z^{2}}\mathrm{d}z.$$ But how to solve further? Please help me out.
Best Answer
The integrand $e^{z^2}$ does not have an elementary antiderivative, but the imaginary error function, a close cousin of the more familiar error function, $\operatorname{erf}$, is purpose-built to express integrals like ours: $$\operatorname{erfi} y := \frac{2}{\sqrt\pi} \int_0^y e^{t^2} \,dt = -i \operatorname{erf} (i y) .$$ Your substitutions amount to $z = \sqrt{1 + \ln x}$, so back-substituting gives that in terms of $\operatorname{erfi}$ the original integral is $$\boxed{\frac{\sqrt\pi}{e} \operatorname{erfi}\sqrt{1 + \ln x} + C} .$$