Mathematics

Solve:
$$\displaystyle \int \dfrac{1 + x + \sqrt{x + x^2}}{\sqrt{x} + \sqrt{1 + x}}dx$$ is equal to


ANSWER

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


SOLUTION
$$\displaystyle\int \dfrac{(1+x)+\sqrt{x+x^2}}{\sqrt{x}+\sqrt{1+x}}dx$$
$$=\displaystyle\int \dfrac{(1+x)+\sqrt{x(1+x)}}{\sqrt{x}+\sqrt{1+x}}\times \dfrac{(\sqrt{x}-\sqrt{1+x})}{(\sqrt{x}-\sqrt{1+x})}dx$$
$$=\displaystyle\int \dfrac{(1+x)+\sqrt{x}\sqrt{(1+x)}}{(x-(1+x))}(\sqrt{x}-\sqrt{1+x})dx$$
$$=\displaystyle\int \dfrac{\sqrt{x}(1+x)-\sqrt{1+x}(1+x)+x\sqrt{1+x}-\sqrt{x}(1+x)}{(-1)}dx$$
$$=\displaystyle\int \dfrac{\sqrt{1+x}(x-1-x)}{(-1)}dx$$
$$=\displaystyle\int \sqrt{1+x}dx$$
$$=\dfrac{2(x+1)^{\dfrac{3}{2}}}{3}+C$$
$$\therefore \displaystyle\int\dfrac{(1+x)+\sqrt{x+x^2}}{\sqrt{x}+\sqrt{1+x}}dx=\dfrac{2(x+1)^{\dfrac{3}{2}}}{3}+C$$.
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Single Correct Medium Published on 17th 09, 2020
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