Question:
Evaluate $\int \frac{1+x^{2}}{\sqrt{1+x^{2}}} d x$
Solution:
$y=\int \sqrt{1+x^{2}} d x$
Use formula $\sqrt{a^{2}+x^{2}}=\frac{x}{2} \sqrt{x^{2}+a^{2}}+\frac{a^{2}}{2} \ln \left(x+\sqrt{x^{2}+a^{2}}\right)$
$y=\frac{x}{2} \sqrt{x^{2}+1}+\frac{1}{2} \ln \left(x+\sqrt{x^{2}+1}\right)+c$
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