Question

plzzz....solve it urgent​

Answer 1

Answer:

here is the explanation for the given question

Answer 2

Step-by-step explanation:

$\sf \int \frac{1+x+\sqrt{x+{x}^{2}}}{\sqrt{x}+\sqrt{1+x}} \:dx \\ \\ \sf \longrightarrow \int \frac{(1+x+\sqrt{x+{x}^{2}})(\sqrt{x}+\sqrt{1+x})}{(\sqrt{x}+\sqrt{1+x})(\sqrt{x}-\sqrt{1+x})}\:dx\\ \\ \sf \longrightarrow \int \frac{(1+x+\sqrt{x+{x}^{2}})(\sqrt{x}+\sqrt{1+x})}{x-1-x} \:dx \\ \\ \sf \longrightarrow - \int (1+x+\sqrt{x+{x}^{2}})(\sqrt{x}-\sqrt{1+x}) \:dx \\ \\ \sf \longrightarrow - \int \sqrt{x} + x\sqrt{x} - (1+x)\sqrt{1+x} +\sqrt{x}\sqrt{x+{x}^{2}} - \sqrt{x+{x}^{2}}\sqrt{1+x} \:dx \\ \\ \sf \longrightarrow - \int \sqrt{x} + {x}^{3/2} -{(1+x)}^{3/2}+ x\sqrt{1+x} -\sqrt{x}(1+x) \:dx \\ \\ \sf \longrightarrow - \int \sqrt{x} + {x}^{3/2} - {(1+x)}^{3/2} +x\sqrt{1+x} - \sqrt{x} - {x}^{3/2} \: dx \\ \\ \sf \longrightarrow </p><p> \int {(1+x)}^{3/2} - x\sqrt{1+x} \: dx \\ \\ \sf \longrightarrow \int(\sqrt{1+x})(1+x + x) \: dx \\ \\ \sf \longrightarrow \int (\sqrt{1+x})(1+2x) \: dx \\ \\ \sf Integrating \: using \: by \: parts \\ \\ \sf \longrightarrow - (1+2x)\int \sqrt{1+x}\:dx + 2\int \sqrt{1+x} \: dx \\ \\ \sf 2×\frac{2}{3}{(1+x)}^{3/2} - \frac{2}{3}(1+2x){(1+x)}^{3/2} +c \\ \\ \sf \longrightarrow \frac{2}{3}{(1+x)}^{3/2}(2+1+2x)+c\\ \\ \sf \longrightarrow \frac{2}{3}{(1+x)}^{3/2}(3+2x)+c$

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