Question

Evaluate the integral \(\int {\frac{x}{{\sqrt {1 + {x^2}} }}} dx\)

Short answer

\(\int {\frac{x}{{\sqrt {1 + {x^2}} }}} dx = \sqrt {1 + {x^2}} + c\)

Step-by-step solution

Simplify the question

\(\int {\frac{x}{{\sqrt {1 + {x^2}} }}} dx\)

Let

\(\begin{aligned}{l}u &= 1 + {x^2}\\du = 2xdx\\dx &= \frac{{du}}{{2x}}\end{aligned}\)

\(\int {\frac{x}{{\sqrt {1 + {x^2}} }}dx = \int {\frac{x}{{\sqrt u }}\frac{{du}}{{2x}}} } \)

Find the integral.

\(\begin{aligned}{l}\int {\frac{x}{{\sqrt {1 + {x^2}} }}dx = \int {\frac{x}{{\sqrt u }}\frac{{du}}{{2x}}} } \\ = \frac{1}{2}\int {\frac{2}{1}{u^{ - \frac{1}{2} + 1}}du} \end{aligned}\)

By adding \(1\) to the power we are making the power positive.

\(\begin{aligned}{l} &= {u^{\frac{1}{2}}}\\ &= \sqrt {1 + {x^2}} + c\end{aligned}\)

Hence, \(\int {\frac{x}{{\sqrt {1 + {x^2}} }}dx = \sqrt {1 + {x^2}} + c} \)