$\int \ln(1+\sqrt{x})\,dx$ any ideas?

Question

$\int \ln(1+\sqrt{x})dx$, any ideas ?

I tried the substitution : $u = 1+ \sqrt{x}$ but it doesn't sem to work

EDIT : Let $u = 1+\sqrt{x}$ then : $du = \frac{1}{2\sqrt{x}}dx \Leftrightarrow dx = du \cdot 2(u-1)$ Hence we have : $\int \ln(1+\sqrt{x})dx = \int \ln(u) \cdot 2(u-1) du$ But this form doesn't seem to help ?

Answer 1

You start with $$\int \ln(1+\sqrt{x}) \, dx$$ then use the substitution $x=u^2$ so that you have $$\int 2u\ln(u+1) \, du$$ Then use integration by parts to get $$u^2\ln(u+1)-\int \frac{u^2}{u+1} \, du$$ Split up the $\frac{u^2}{u+1}$ using partial fractions $$u^2\ln(u+1)-\int (u-1+\frac{1}{u+1}) \, du$$ This is now a pretty straightforward integral. Continue on to get $$u^2\ln(u+1)-\frac{1}{2}u^2+u-ln(u+1)$$ $$(u^2-1)\ln(u+1)-\frac{1}{2}u^2+u+C$$ Then substitute $u=\sqrt{x}$: $$(x-1)\ln(\sqrt{x}+1)-\frac{1}{2}x+\sqrt{x}+C$$ Which should be the answer.

Answer 2

We can do the first phase by parts: $$\int \ln(1+\sqrt x)\ \mathrm dx = x \ln(1+\sqrt x) - \frac 12 \int \frac{\sqrt x}{1 + \sqrt x}\ \mathrm dx$$

Let $u = \sqrt x$; then $x = u^2$ and $\mathrm dx = 2u\ \mathrm du$, so: $$\int \frac{\sqrt x}{1+\sqrt x}\ \mathrm dx = \int \frac{2u^2}{1 + u}\ \mathrm du = \int 2\left(u - \frac u{1+u}\right)\mathrm du = \int 2\left(u - 1 + \frac 1{u+1}\right)\mathrm du = u^2 - 2u + 2\ln(u+1)$$

So, the final formula becomes $$\int \ln(1 + \sqrt x)\mathrm dx = x\ln(1+\sqrt x) - \frac x2 + \sqrt x - \ln(1+\sqrt x) = (x-1)\ln(1+\sqrt x) + \sqrt x - \frac x2 + C$$

Answer 3

\begin{align} u & = 1 + \sqrt x \\[10pt] u-1 & = \sqrt x \\[10pt] (u-1)^2 & = x \\[10pt] 2(u-1)\,du & = dx \\[10pt] \int \ln(1+\sqrt x) \, dx & = \int (\ln u) \Big(2(u-1)\, du\Big) \\[10pt] = \int w \, dv & = wv - \int v\,dw = (\ln u)(u-1)^2 - \int (u-1)^2 \left( \frac{du} u \right) \\[10pt] & = (\ln u)(u-1)^2 - \int\left( u - 2 + \frac 1 u \right) \, du = \cdots\cdots \end{align}