evaluate the integral. (remember the constant of integration.)\n$$ int \frac { x } { sqrt { x ^ { 2 } - 6 }…

evaluate the integral. (remember the constant of integration.)\n$$ int \frac { x } { sqrt { x ^ { 2 } - 6 } } d x $$\n$$ 6 left( \frac { sqrt { 6 } } { sqrt { x ^ { 2 } - 6 } } \right) + c $$

evaluate the integral. (remember the constant of integration.)\n$$ int \frac { x } { sqrt { x ^ { 2 } - 6 } } d x $$\n$$ 6 left( \frac { sqrt { 6 } } { sqrt { x ^ { 2 } - 6 } } \right) + c $$

Answer

Explanation:

Step1: Use substitution

Let ( u = x^{2}-6 ), then ( du = 2x dx), and (x dx=\frac{1}{2}du). The integral (\int\frac{x}{\sqrt{x^{2}-6}}dx) becomes (\frac{1}{2}\int u^{-\frac{1}{2}}du).

Step2: Integrate using power rule

The power - rule for integration (\int x^{n}dx=\frac{x^{n + 1}}{n+1}+C) ((n\neq - 1)). For (\frac{1}{2}\int u^{-\frac{1}{2}}du), we have (\frac{1}{2}\times\frac{u^{-\frac{1}{2}+1}}{-\frac{1}{2}+1}+C). Simplify the exponent and coefficient: (\frac{1}{2}\times\frac{u^{\frac{1}{2}}}{\frac{1}{2}}+C=\sqrt{u}+C).

Step3: Substitute back

Substitute (u = x^{2}-6) back into the expression. We get (\sqrt{x^{2}-6}+C).

Answer:

(\sqrt{x^{2}-6}+C)