Answer on Question #51420 – Math – Integral Calculus
Question
∫ x 4 x 3 + x − 2 d x = ? \int \frac{x^4}{x^3 + x^{-2}} \, dx = ? ∫ x 3 + x − 2 x 4 d x = ? Solution
∫ x 4 x 3 + x − 2 d x = ∫ x 4 x 3 + 1 x 2 d x = ∫ x 4 x 2 x 5 + 1 d x = ∫ x 6 x 5 + 1 d x = ∫ ( x − x x 5 + 1 ) d x = x 2 2 − ∫ x d x x 5 + 1 = x 2 2 − ∫ x d x ( x + 1 ) ( x 4 − x 3 + x 2 − x + 1 ) = x 2 2 − ∫ x d x ( x + 1 ) ( x 2 − 5 + 1 2 x + 1 ) ( x 2 + 5 − 1 2 x + 1 ) = \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} ∫ x 3 + x − 2 x 4 d x = ∫ x 3 + x 2 1 x 4 d x = ∫ x 4 x 5 + 1 x 2 d x = ∫ x 5 + 1 x 6 d x = ∫ ( x − x 5 + 1 x ) d x = 2 x 2 − ∫ x 5 + 1 x d x = 2 x 2 − ∫ ( x + 1 ) ( x 4 − x 3 + x 2 − x + 1 ) x d x = 2 x 2 − ∫ ( x + 1 ) ( x 2 − 2 5 + 1 x + 1 ) ( x 2 + 2 5 − 1 x + 1 ) x d x = = x 2 2 − ∫ ( − 1 5 ( x + 1 ) + x ( 5 + 1 ) + ( 1 − 5 ) 5 ( 2 x 2 + ( 5 − 1 ) x + 2 ) + x ( 1 − 5 ) + ( 5 + 1 ) 5 ( 2 x 2 − ( 5 + 1 ) x + 2 ) ) d x = = x 2 2 + 1 5 ∫ d x x + 1 − 5 + 1 5 ∫ x d x 2 x 2 + ( 5 − 1 ) x + 2 − 1 − 5 5 ∫ d x 2 x 2 − ( 5 + 1 ) x + 2 − 1 − 5 5 ∫ x d x 2 x 2 + ( 5 − 1 ) x + 2 − − 5 + 1 5 ∫ d x 2 x 2 − ( 5 + 1 ) x + 2 = x 2 2 + 1 5 ln ∣ x + 1 ∣ − 5 + 1 20 ln ∣ 2 x 2 + ( 5 − 1 ) x + 2 ∣ + + 5 − 1 20 ln ∣ − 2 x 2 + ( 5 + 1 ) x − 2 ∣ + 2 5 arctan ( 4 x + 5 − 1 10 + 2 5 ) 10 + 2 5 + 2 5 arctan ( 5 + 1 − 4 x 10 − 2 5 ) 10 − 2 5 + c . \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} = 2 x 2 − ∫ ( − 5 ( x + 1 ) 1 + 5 ( 2 x 2 + ( 5 − 1 ) x + 2 ) x ( 5 + 1 ) + ( 1 − 5 ) + 5 ( 2 x 2 − ( 5 + 1 ) x + 2 ) x ( 1 − 5 ) + ( 5 + 1 ) ) d x = = 2 x 2 + 5 1 ∫ x + 1 d x − 5 5 + 1 ∫ 2 x 2 + ( 5 − 1 ) x + 2 x d x − 5 1 − 5 ∫ 2 x 2 − ( 5 + 1 ) x + 2 d x − 5 1 − 5 ∫ 2 x 2 + ( 5 − 1 ) x + 2 x d x − − 5 5 + 1 ∫ 2 x 2 − ( 5 + 1 ) x + 2 d x = 2 x 2 + 5 1 ln ∣ x + 1∣ − 20 5 + 1 ln ∣2 x 2 + ( 5 − 1 ) x + 2∣ + + 20 5 − 1 ln ∣ ∣ − 2 x 2 + ( 5 + 1 ) x − 2 ∣ ∣ + 5 2 10 + 2 5 arctan ( 10 + 2 5 4 x + 5 − 1 ) + 5 2 10 − 2 5 arctan ( 10 − 2 5 5 + 1 − 4 x ) + c . Answer: x 2 2 + 1 5 ln ∣ x + 1 ∣ − 5 + 1 20 ln ∣ 2 x 2 + ( 5 − 1 ) x + 2 ∣ + + 5 − 1 20 ln ∣ − 2 x 2 + ( 5 + 1 ) x − 2 ∣ + 2 5 arctan ( 4 x + 5 − 1 10 + 2 5 ) 10 + 2 5 + 2 5 arctan ( 5 + 1 − 4 x 10 − 2 5 ) 10 − 2 5 + c . \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} Answer: 2 x 2 + 5 1 ln ∣ x + 1∣ − 20 5 + 1 ln ∣ ∣ 2 x 2 + ( 5 − 1 ) x + 2 ∣ ∣ + + 20 5 − 1 ln ∣ ∣ − 2 x 2 + ( 5 + 1 ) x − 2 ∣ ∣ + 5 2 10 + 2 5 arctan ( 10 + 2 5 4 x + 5 − 1 ) + 5 2 10 − 2 5 arctan ( 10 − 2 5 5 + 1 − 4 x ) + c .
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#339914 on Dec 2023
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