Question

Use any method to evaluate the integrals in Exercises 55–66.∫ x / (x + √(x² + 2)) dx

Accepted Answer

Start by examining the integral: $$\int \frac{x}{x + \sqrt{x$$^{2}$$ + 2}} \, dx. $$Notice the expression in the denominator involves $$x$$ and $$\sqrt{x$$^{2}$$ + 2}$$, which suggests a substitution related to the square root term. Multiply the numerator and denominator by the conjugate of the denominator to simplify the integrand. The conjugate of $$x + \sqrt{x$$^{2}$$ + 2} is$$ $$\sqrt{x$$^{2}$$ + 2} - x. So$$, multiply numerator and denominator by $$\sqrt{x$$^{2}$$ + 2} - x$$: $$\frac{x}{x + \sqrt{x^{2} + 2}} 	imes \frac{\sqrt{x^{2} + 2} - x}{\sqrt{x^{2} + 2} - x} = \frac{x(\sqrt{x^{2} + 2} - x)}{(x + \sqrt{x^{2} + 2})(\sqrt{x^{2} + 2} - x)}.$$ Simplify the denominator using the difference of squares formula: $$(a + b)(a - b) = a$$^{2}$$ - b$$^{2}$$. Here, a$$ = \sqrt{$$x^{2} + 2$$}$$ and b = x$, so the denominator becomes $$(\sqrt{$$x^{2} + 2$$}$$)^{2}$$ - $$x^{2}$$ = $$(x^{2} + 2$$) - $$x^{2} = 2$. Rewrite the integral as $$\int \frac{x \sqrt{$$x^{2} + 2$$} - $$x^{2}$$}{2} \, dx = \frac{1}{2} \int (x \sqrt{$$x^{2} + 2$$} - $$x^{2}$$) \, dx$$. Now, split the integral into two separate integrals: $$\frac{1}{2} \int x \sqrt{x^{2} + 2} \, dx - \frac{1}{2} \int x^{2} \, dx.$$

Use any method to evaluate the integrals in Exercises 55–66.
∫ x / (x + √(x² + 2)) dx

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