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

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(B) Given integral is $I = \int \frac{1+x+\sqrt{x+x^2}}{\sqrt{x}+\sqrt{1+x}} d x$. We can rewrite the numerator as: $1+x+\sqrt{x(1+x)} = \sqrt{1+x} \c

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

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(B) Given integral is $I = \int \frac{1+x+\sqrt{x+x^2}}{\sqrt{x}+\sqrt{1+x}} d x$. We can rewrite the numerator as: $1+x+\sqrt{x(1+x)} = \sqrt{1+x} \cdot \sqrt{1+x} + \sqrt{x} \cdot \sqrt{1+x}$. Factoring out $\sqrt{1+x}$, we get: $\sqrt{1+x}(\sqrt{1+x} + \sqrt{x})$. Substituting this back into the integral: $I = \int \frac{\sqrt{1+x}(\sqrt{1+x} + \sqrt{x})}{\sqrt{x}+\sqrt{1+x}} d x$. The term $(\sqrt{x}+\sqrt{1+x})$ cancels out from the numerator and denominator. Thus, $I = \int \sqrt{1+x} d x$. Using the power rule $\int u^n du = \frac{u^{n+1}}{n+1} + C$, we get: $I = \frac{(1+x)^{3/2}}{3/2} + C = \frac{2}{3}(1+x)^{3/2} + C$.

$\int \frac{1+x+\sqrt{x+x^2}}{\sqrt{x}+\sqrt{1+x}} d x$ is equal to

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