Question:

Let $I = \int \frac{\ln \left(x+\sqrt{1+x^2}\right)}{\sqrt{1+x^2}} d x$. Let $u = \ln \left(x+\sqrt{1+x^2}\right)$, then $du = \frac{1}{\sqrt{1+x^2}} dx$. Let $dv = \frac{1}{\sqrt{1+x^2}} dx$, then $v = \ln \left(x+\sqrt{1+x^2}\right)$. This is not correct. Let $I = \int u \cdot dv$. Let $u = \ln \left(x+\sqrt{1+x^2}\right)$ and $dv = \frac{1}{\sqrt{1+x^2}} dx$. Then $du = \frac{1}{\sqrt{1+x^2}} dx$ and $v = \ln \left(x+\sqrt{1+x^2}\right)$. This leads to $I = u^2 - I$, so $2I = u^2$, $I = u^2/2$. This does not match the options. Let's re-evaluate. Let $u = \ln \left(x+\sqrt{1+x^2}\right)$, then $du = \frac{1}{\sqrt{1+x^2}} dx$. The integral is $\int u du = u^2/2 + c = \frac{1}{2} [\ln(x+\sqrt{1+x^2})]^2 + c$. None of the options match this. Let's try integration by parts again: $\int u dv = uv - \int v du$. Let $u = \ln(x+\sqrt{1+x^2})$, $dv = \frac{1}{\sqrt{1+x^2}} dx$. Then $du = \frac{1}{\sqrt{1+x^2}} dx$. This is not working. Let's differentiate option (A): $\frac{d}{dx} [\sqrt{1+x^2} \ln(x+\sqrt{1+x^2}) - x] = \frac{x}{\sqrt{1+x^2}} \ln(x+\sqrt{1+x^2}) + \sqrt{1+x^2} \cdot \frac{1}{\sqrt{1+x^2}} - 1 = \frac{x}{\sqrt{1+x^2}} \ln(x+\sqrt{1+x^2})$. This is not the integrand. Let's re-read the question. Maybe it is $\int \ln(x+\sqrt{1+x^2}) dx$. If $I = \int \ln(x+\sqrt{1+x^2}) dx$, then $I = x \ln(x+\sqrt{1+x^2}) - \int x \cdot \frac{1}{\sqrt{1+x^2}} dx = x \ln(x+\sqrt{1+x^2}) - \sqrt{1+x^2} + c$. This matches option (A) if the denominator $\sqrt{1+x^2}$ was not there.