$$\int \frac{1+x+\sqrt{x+x^2}}{\sqrt{x}+\sqrt{1+x}} d x$$ is equal to
- A$$\frac{1}{2}(\sqrt{1+x})+c$$
- B$$\sqrt{1+x}+c$$
- C$$2(1+x)^{3 / 2}+c$$
- D$$\frac{2}{3}(1+x)^{3 / 2}+c$$
✓ Correct answer: D
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$$\int \frac{1+x+\sqrt{x+x^2}}{\sqrt{x}+\sqrt{1+x}} d x$$ is equal to
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Question ID #140194 · NewNcert