I want to evaluate the integral
$$\int_0^1\sqrt{\frac1x+1}\,dx.$$
Letting $u=\sqrt{\frac1x+1}$, the integral becomes
$$\int_{\sqrt 2}^\infty\frac{2u^2}{(u^2-1)^2}\,du,$$
which according to mathematica, equals $\sqrt2+\frac12\log(3+2\sqrt2)$. But using partial fractions, the antiderivative of $\frac{2u^2}{(u^2-1)^2}$ is
$$\int\frac{2u^2}{(u^2-1)^2}\,du=\frac{u}{1-u^2}+\frac12\log\Big(\frac2{1+u}-1\Big),$$
which I can't really make sense of as $u\to\infty$ to apply FTC (or when $u=\sqrt2$ for that matter). Is there an easier way to go about this integral? Or do we necessarily need to involve complex analysis because of log branches?
The improper integral $\int_0^1\sqrt{\frac1x+1}\,dx$
calculusintegration
Best Answer
As $x \ge 0$ an alternative substitution is let $x=u^2$. Then $$ 2u\,du=dx $$ and the integral becomes $$ 2\int_0^1 \sqrt{1+u^2}\,du $$ which probably looks much more familiar. Interestingly this leads to a different form for the result $$ \sqrt2+\log(\sqrt2+1) $$ This is ok, this tells us, comparing the $\log$ terms $$ \sqrt{3+2\sqrt{2}}=\sqrt2+1 $$ which of course it is (square both sides).