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\( \int \frac { 4 x } { x ^ { 5 } - 2 } d x \)

Ask by Love Ellis. in South Africa
Mar 17,2025

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Answer

The integral \( \int \frac{4x}{x^5 - 2} \, dx \) equals \( \frac{4}{5} \ln |x^5 - 2| + C \).

Solution

It seems that the integral \( \int \frac{4x}{x^5 - 2} \, dx \) cannot be directly computed using the available tools. However, I can guide you through the process of solving it step by step. ### Step 1: Identify the Integral We need to evaluate the integral: \[ I = \int \frac{4x}{x^5 - 2} \, dx \] ### Step 2: Use Substitution A common technique for integrals of this form is to use substitution. Let's set: \[ u = x^5 - 2 \] Then, we differentiate \(u\) with respect to \(x\): \[ \frac{du}{dx} = 5x^4 \quad \Rightarrow \quad du = 5x^4 \, dx \quad \Rightarrow \quad dx = \frac{du}{5x^4} \] ### Step 3: Rewrite the Integral Now, we can express \(x\) in terms of \(u\): \[ x^4 = (u + 2)^{4/5} \] Substituting \(dx\) and \(u\) into the integral gives: \[ I = \int \frac{4x}{u} \cdot \frac{du}{5x^4} \] This simplifies to: \[ I = \frac{4}{5} \int \frac{1}{u} \, du \] ### Step 4: Integrate The integral of \(\frac{1}{u}\) is: \[ \int \frac{1}{u} \, du = \ln |u| + C \] Thus, we have: \[ I = \frac{4}{5} \ln |u| + C \] ### Step 5: Substitute Back Now, substitute back \(u = x^5 - 2\): \[ I = \frac{4}{5} \ln |x^5 - 2| + C \] ### Final Answer The final result of the integral is: \[ \int \frac{4x}{x^5 - 2} \, dx = \frac{4}{5} \ln |x^5 - 2| + C \] This is the solution to the integral. If you have any further questions or need additional assistance, feel free to ask!

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Beyond the Answer

To tackle the integral \( \int \frac { 4 x } { x ^ { 5 } - 2 } d x \), first, we can use substitution to simplify our work. Notice that if we let \( u = x^5 - 2 \), then \( du = 5x^4 \, dx \) or \( dx = \frac{du}{5x^4} \). This substitution will help transform the integral. However, as we simplify, we must also consider the expression for \( x \) in terms of \( u \) to adjust the integrand accordingly. Another approach is to recognize that the numerator can be rearranged to fit the format of the derivative of the denominator, leading to a logarithmic function upon integration. Don’t forget to check common mistakes, such as forgetting the limits if you’re evaluating a definite integral or miscalculating variable substitutions! To deepen your understanding, exploring integration techniques like substitution, integration by parts, or partial fractions in calculus can be quite beneficial. There’s a wealth of resources available, including classic textbooks like "Calculus" by James Stewart or online platforms like Khan Academy, which break down the steps engagingly. Happy integrating!

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