Math, asked by dnyanudhande, 5 months ago

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Answered by Thatsomeone
5

Step-by-step explanation:

[tex]\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

- \int {(1+x)}^{3/2} + x\sqrt{1+x} \: dx \\ \\ \sf \longrightarrow - \int(\sqrt{1+x})(1+2x) \: dx \\ \\ \sf \longrightarrow \int \sqrt{1+x} \: dx \\ \\ \sf \l

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