Question

∫ {(x)^4} / [(x)^3 + (x)^(-2)] dx =? show process please.?

Answer 1

Answer on Question #51420 – Math – Integral Calculus

Question

$$\int \frac{x^4}{x^3 + x^{-2}} \, dx = ?$$

Solution

$$\begin{aligned} & \int \frac{x^4}{x^3 + x^{-2}} \, dx = \int \frac{x^4}{x^3 + \frac{1}{x^2}} \, dx = \int x^4 \frac{x^2}{x^5 + 1} \, dx = \int \frac{x^6}{x^5 + 1} \, dx = \int \left(x - \frac{x}{x^5 + 1}\right) dx \\ & = \frac{x^2}{2} - \int \frac{x \, dx}{x^5 + 1} = \frac{x^2}{2} - \int \frac{x \, dx}{(x + 1)(x^4 - x^3 + x^2 - x + 1)} \\ & = \frac{x^2}{2} - \int \frac{x \, dx}{(x + 1)\left(x^2 - \frac{\sqrt{5} + 1}{2}x + 1\right)\left(x^2 + \frac{\sqrt{5} - 1}{2}x + 1\right)} = \\ \end{aligned}$$ $$\begin{aligned} & = \frac{x^2}{2} - \int \left(- \frac{1}{5(x + 1)} + \frac{x(\sqrt{5} + 1) + (1 - \sqrt{5})}{5(2x^2 + (\sqrt{5} - 1)x + 2)} + \frac{x(1 - \sqrt{5}) + (\sqrt{5} + 1)}{5(2x^2 - (\sqrt{5} + 1)x + 2)}\right) dx = \\ & = \frac{x^2}{2} + \frac{1}{5} \int \frac{dx}{x + 1} - \frac{\sqrt{5} + 1}{5} \int \frac{x \, dx}{2x^2 + (\sqrt{5} - 1)x + 2} - \frac{1 - \sqrt{5}}{5} \int \frac{dx}{2x^2 - (\sqrt{5} + 1)x + 2} - \frac{1 - \sqrt{5}}{5} \int \frac{x \, dx}{2x^2 + (\sqrt{5} - 1)x + 2} - \\ & - \frac{\sqrt{5} + 1}{5} \int \frac{dx}{2x^2 - (\sqrt{5} + 1)x + 2} = \frac{x^2}{2} + \frac{1}{5} \ln |x + 1| - \frac{\sqrt{5} + 1}{20} \ln |2x^2 + (\sqrt{5} - 1)x + 2| + \\ & + \frac{\sqrt{5} - 1}{20} \ln \left| -2x^2 + (\sqrt{5} + 1)x - 2 \right| + \frac{2}{\sqrt{5}} \frac{\arctan\left(\frac{4x + \sqrt{5} - 1}{\sqrt{10 + 2\sqrt{5}}}\right)}{\sqrt{10 + 2\sqrt{5}}} + \frac{2}{\sqrt{5}} \frac{\arctan\left(\frac{\sqrt{5} + 1 - 4x}{\sqrt{10 - 2\sqrt{5}}}\right)}{\sqrt{10 - 2\sqrt{5}}} + c. \end{aligned}$$ $$\begin{aligned} & \text{Answer: } \frac{x^2}{2} + \frac{1}{5} \ln |x + 1| - \frac{\sqrt{5} + 1}{20} \ln \left| 2x^2 + (\sqrt{5} - 1)x + 2 \right| + \\ & + \frac{\sqrt{5} - 1}{20} \ln \left| -2x^2 + (\sqrt{5} + 1)x - 2 \right| + \frac{2}{\sqrt{5}} \frac{\arctan\left(\frac{4x + \sqrt{5} - 1}{\sqrt{10 + 2\sqrt{5}}}\right)}{\sqrt{10 + 2\sqrt{5}}} + \frac{2}{\sqrt{5}} \frac{\arctan\left(\frac{\sqrt{5} + 1 - 4x}{\sqrt{10 - 2\sqrt{5}}}\right)}{\sqrt{10 - 2\sqrt{5}}} + c. \end{aligned}$$

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