int x cdot frac{ln(x + sqrt{1 + x^2})}{sqrt{1 | vedclass

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(A) Let $I = \int \frac{x}{\sqrt{1 + x^2}} \ln(x + \sqrt{1 + x^2}) \, dx$. Using Integration by Parts,let $u = \ln(x + \sqrt{1 + x^2})$ and $dv = \fra

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$\sqrt{1 + x^2} \ln(x + \sqrt{1 + x^2}) - x + c$

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(A) Let $I = \int \frac{x}{\sqrt{1 + x^2}} \ln(x + \sqrt{1 + x^2}) \, dx$. Using Integration by Parts,let $u = \ln(x + \sqrt{1 + x^2})$ and $dv = \frac{x}{\sqrt{1 + x^2}} \, dx$. Then $du = \frac{1}{x + \sqrt{1 + x^2}} \cdot (1 + \frac{x}{\sqrt{1 + x^2}}) \, dx = \frac{1}{x + \sqrt{1 + x^2}} \cdot \frac{\sqrt{1 + x^2} + x}{\sqrt{1 + x^2}} \, dx = \frac{1}{\sqrt{1 + x^2}} \, dx$. And $v = \int \frac{x}{\sqrt{1 + x^2}} \, dx = \sqrt{1 + x^2}$. Applying the formula $\int u \, dv = uv - \int v \, du$: $I = \sqrt{1 + x^2} \ln(x + \sqrt{1 + x^2}) - \int \sqrt{1 + x^2} \cdot \frac{1}{\sqrt{1 + x^2}} \, dx$. $I = \sqrt{1 + x^2} \ln(x + \sqrt{1 + x^2}) - \int 1 \, dx$. $I = \sqrt{1 + x^2} \ln(x + \sqrt{1 + x^2}) - x + c$.

$\int x \cdot \frac{\ln(x + \sqrt{1 + x^2})}{\sqrt{1 + x^2}} \, dx$ equals :

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