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

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(B) Given integral is $I = \int \frac{1 + x + \sqrt{x(1 + x)}}{\sqrt{x} + \sqrt{1 + x}} dx$. We can rewrite the numerator as: $1 + x + \sqrt{x} \cdot

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

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(B) Given integral is $I = \int \frac{1 + x + \sqrt{x(1 + x)}}{\sqrt{x} + \sqrt{1 + x}} dx$. We can rewrite the numerator as: $1 + x + \sqrt{x} \cdot \sqrt{1 + x} = \sqrt{1 + x} \cdot \sqrt{1 + x} + \sqrt{x} \cdot \sqrt{1 + x}$. Factoring out $\sqrt{1 + x}$,we get: $\sqrt{1 + x} (\sqrt{1 + x} + \sqrt{x})$. Substituting this back into the integral: $I = \int \frac{\sqrt{1 + x} (\sqrt{1 + x} + \sqrt{x})}{\sqrt{x} + \sqrt{1 + x}} dx$. Canceling the common term $(\sqrt{x} + \sqrt{1 + x})$: $I = \int \sqrt{1 + x} dx = \int (1 + x)^{1/2} dx$. Using the power rule $\int u^n du = \frac{u^{n+1}}{n+1} + c$: $I = \frac{(1 + x)^{3/2}}{3/2} + c = \frac{2}{3} (1 + x)^{3/2} + c$.

$\int \frac{1 + x + \sqrt{x + x^2}}{\sqrt{x} + \sqrt{1 + x}} dx =$

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