For x>0 ∫(1+x+x21+x+121+x+x2−1(1+x)1+x+x2) dx

∫2(1+x+x2)+1+x−22(1+x)1+x+x2 dx⇒∫2x(1+x)+(1+x)2(1+x)1+x+x2 dx=∫2x+121+x+x2dx=1+x+x2+C

For x>0 ∫(1+x+x21+x+121+x+x2−1(1+x)1+x+x2) dx

∫2(1+x+x2)+1+x−22(1+x)1+x+x2 dx⇒∫2x(1+x)+(1+x)2(1+x)1+x+x2 dx=∫2x+121+x+x2dx=1+x+x2+C

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For $x > 0 \int (\frac{\sqrt{1 + x + x^{2}}}{1 + x} + \frac{1}{2 \sqrt{1 + x + x^{2}}} - \frac{1}{(1 + x) \sqrt{1 + x + x^{2}}}) d x$

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$\frac{1}{\sqrt{1 + x + x^{2}}} + C$

$\sqrt{1 + x} + c$

$\frac{1}{\sqrt{1 + x}} + C$

$\sqrt{1 + x + x^{2}} + C$

answer is D.

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Detailed Solution

$\int \frac{2 (1 + x + x^{2}) + 1 + x - 2}{2 (1 + x) \sqrt{1 + x + x^{2}}} d x \Rightarrow \int \frac{2 x (1 + x) + (1 + x)}{2 (1 + x) \sqrt{1 + x + x^{2}}} d x = \int \frac{2 x + 1}{2 \sqrt{1 + x + x^{2}}} d x = \sqrt{1 + x + x^{2}} + C$

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For x>0 ∫(1+x+x21+x+121+x+x2−1(1+x)1+x+x2) dx

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For $x > 0 \int (\frac{\sqrt{1 + x + x^{2}}}{1 + x} + \frac{1}{2 \sqrt{1 + x + x^{2}}} - \frac{1}{(1 + x) \sqrt{1 + x + x^{2}}}) d x$