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\( \int \sqrt{x^{2}+2 x+5} d x \) is equal to
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Verified Answer
The correct answer is:
\( \frac{1}{2}(x+1) \sqrt{x^{2}+2 x+5}+2 \log \left|x+1+\sqrt{x^{2}+2 x+5}\right|+C \)
Given that, \( \int \sqrt{x^{2}+2 x+5} d x \)
\[
=\int \sqrt{x^{2}+2 x+1+4} d x
\]
\[
=\int \sqrt{(x+1)^{2}+2^{2}} d x
\]
We know that,
\( \int \sqrt{x^{2}+a^{2}} d x=\frac{x}{2} \sqrt{x^{2}+a^{2}}+2 \log \left|x+\sqrt{x^{2}+a^{2}}\right|+C \)
So, \( \int \sqrt{x^{2}+2 x+5} d x \)
\( =\frac{(x+1)}{2} \sqrt{x^{2}+2 x+5}+2 \log \left|(x+1)+\sqrt{x^{2}+2 x+5}\right|+C \)
\[
=\int \sqrt{x^{2}+2 x+1+4} d x
\]
\[
=\int \sqrt{(x+1)^{2}+2^{2}} d x
\]
We know that,
\( \int \sqrt{x^{2}+a^{2}} d x=\frac{x}{2} \sqrt{x^{2}+a^{2}}+2 \log \left|x+\sqrt{x^{2}+a^{2}}\right|+C \)
So, \( \int \sqrt{x^{2}+2 x+5} d x \)
\( =\frac{(x+1)}{2} \sqrt{x^{2}+2 x+5}+2 \log \left|(x+1)+\sqrt{x^{2}+2 x+5}\right|+C \)
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