Integration by substitution
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Find \(\int {\dfrac{x}{{\sqrt {1 + {x^2}} }} dx= } \)
\begin{aligned}u =1+x^{2}\qquad\qquad \quad &\\\frac{d u}{d x}=2 x \rightarrow \frac{1}{2} d u= x d x& \\\int \frac{x}{\sqrt{1+x^{2}}} d x=\frac{1}{2} \int \frac{d u}{\sqrt{u}}&= \frac{1}{2} \int u^{-\frac{1}{2}} d d \\&=\frac{1}{2} \times \frac{u^{\frac{1}{2}}}{\frac{1}{2}}+c \\&=\sqrt{1+x^{2}}+c\end{aligned}
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