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

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Question

(B) We are given the integral $I = \int \frac{1+x+\sqrt{x+x^2}}{\sqrt{x}+\sqrt{1+x}} d x$. First,we rewrite the numerator as $(1+x) + \sqrt{x(1+x)}$.

Vedclass · Accepted Answer

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

Vedclass · Answer

(B) We are given the integral $I = \int \frac{1+x+\sqrt{x+x^2}}{\sqrt{x}+\sqrt{1+x}} d x$. First,we rewrite the numerator as $(1+x) + \sqrt{x(1+x)}$. This can be factored as $\sqrt{1+x}(\sqrt{1+x} + \sqrt{x})$. Substituting this back into the integral,we get: $I = \int \frac{\sqrt{1+x}(\sqrt{1+x} + \sqrt{x})}{\sqrt{x} + \sqrt{1+x}} d x$. The term $(\sqrt{1+x} + \sqrt{x})$ cancels out from the numerator and denominator. $I = \int \sqrt{1+x} d x$. Using the power rule for integration $\int u^n du = \frac{u^{n+1}}{n+1} + C$,where $u = 1+x$ and $du = dx$: $I = \frac{(1+x)^{3/2}}{3/2} + C = \frac{2}{3}(1+x)^{3/2} + C$.

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

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