int left( frac{1+x+sqrt{x+x^2}}{sqrt{x}+sqrt{ | vedclass

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(B) Let $I = \int \frac{1+x+\sqrt{x(1+x)}}{\sqrt{x}+\sqrt{1+x}} dx$. Notice that $1+x = (\sqrt{1+x})^2$ and $x = (\sqrt{x})^2$. So,the numerator can b

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$\frac{2}{3}(1+x)^{3/2} + c$

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(B) Let $I = \int \frac{1+x+\sqrt{x(1+x)}}{\sqrt{x}+\sqrt{1+x}} dx$. Notice that $1+x = (\sqrt{1+x})^2$ and $x = (\sqrt{x})^2$. So,the numerator can be written as $(\sqrt{1+x})^2 + (\sqrt{x})^2 + \sqrt{x}\sqrt{1+x}$. This does not simplify directly,so let us rewrite the integrand: $I = \int \frac{(\sqrt{1+x})^2 + \sqrt{x}\sqrt{1+x}}{\sqrt{x}+\sqrt{1+x}} dx$. Factor out $\sqrt{1+x}$ from the numerator: $I = \int \frac{\sqrt{1+x}(\sqrt{1+x} + \sqrt{x})}{\sqrt{x}+\sqrt{1+x}} dx$. Canceling the common term $(\sqrt{x}+\sqrt{1+x})$,we get: $I = \int \sqrt{1+x} dx$. Using the power rule $\int u^n du = \frac{u^{n+1}}{n+1} + c$: $I = \frac{(1+x)^{3/2}}{3/2} + c = \frac{2}{3}(1+x)^{3/2} + c$.

$\int \left( \frac{1+x+\sqrt{x+x^2}}{\sqrt{x}+\sqrt{1+x}} \right) dx =$

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