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$\int \frac{x+1}{\sqrt{1+x^2}} d x=$
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Verified Answer
The correct answer is:
$\sqrt{1+x^2}+\log \left\{x+\sqrt{1+x^2}\right\}+c$
$\int \frac{x+1}{\sqrt{x^2+1}} d x=\int \frac{x}{\sqrt{x^2+1}} d x+\int \frac{1}{\sqrt{x^2+1}} d x$
Put $x^2+1=t \Rightarrow 2 x d x=d t$, then it reduce to $\frac{1}{2} \int \frac{d t}{t^{1 / 2}}+\int \frac{1}{\sqrt{x^2+1}} d x=\frac{1}{2} \cdot 2 \cdot t^{1 / 2}+\log \left(x+\sqrt{x^2+1}\right)+c$
$=\left(x^2+1\right)^{1 / 2}+\log \left(x+\sqrt{x^2+1}\right)+c .$
Put $x^2+1=t \Rightarrow 2 x d x=d t$, then it reduce to $\frac{1}{2} \int \frac{d t}{t^{1 / 2}}+\int \frac{1}{\sqrt{x^2+1}} d x=\frac{1}{2} \cdot 2 \cdot t^{1 / 2}+\log \left(x+\sqrt{x^2+1}\right)+c$
$=\left(x^2+1\right)^{1 / 2}+\log \left(x+\sqrt{x^2+1}\right)+c .$
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