Answer to Question #266487 in Calculus for Faru

Question #266487

Integrate (3-x^2)/(x^2+3)^2


1
Expert's answer
2021-11-16T11:36:38-0500

"\\begin{aligned}\n&\\int \\frac{3-x^{2}}{\\left(x^{2}+3\\right)^{2}} \\mathrm{~d} x \\\\\n&=-\\int \\frac{x^{2}-3}{\\left(x^{2}+3\\right)^{2}} \\mathrm{~d} x \\\\\n&=-\\int\\left(\\frac{x^{2}+3}{\\left(x^{2}+3\\right)^{2}}-\\frac{6}{\\left(x^{2}+3\\right)^{2}}\\right) \\mathrm{d} x \\\\\n&=-\\int\\left(\\frac{1}{x^{2}+3}-\\frac{6}{\\left(x^{2}+3\\right)^{2}}\\right) \\mathrm{d} x \\\\\n&=-\\int \\frac{1}{x^{2}+3} \\mathrm{~d} x-6 \\int \\frac{1}{\\left(x^{2}+3\\right)^{2}} \\mathrm{~d} x \\\\\n&=-\\frac{\\arctan \\left(\\frac{x}{\\sqrt{3}}\\right)}{\\sqrt{3}}+6\\left[\\frac{x}{6\\left(x^{2}+3\\right)}+\\frac{1}{6} \\int \\frac{1}{x^{2}+3} \\mathrm{~d} x\\right] \\\\\n&[\\because \\frac{1}{\\left(\\mathrm{a} x^{2}+\\mathrm{b}\\right)^{\\mathrm{n}}} \\mathrm{d} x=\\frac{2 \\mathrm{n}-3}{2 \\mathrm{~b}(\\mathrm{n}-1)} \\int \\frac{1}{\\left(\\mathrm{a} x^{2}+\\mathrm{b}\\right)^{\\mathrm{n}-1}} \\mathrm{~d} x+\\frac{x}{2 \\mathrm{~b}(\\mathrm{n}-1)\\left(\\mathrm{a} x^{2}+\\mathrm{b}\\right)^{\\mathrm{n}-1}} ]\\\\\n&=-\\frac{\\arctan \\left(\\frac{x}{\\sqrt{3}}\\right)}{\\sqrt{3}}+ \\frac{x}{x^{2}+3}+\\frac{\\arctan \\left(\\frac{x}{\\sqrt{3}}\\right)}{\\sqrt{3}}\n=\\frac{x}{x^{2}+3}\n\\end{aligned}"


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