Questions: ∫ 1/(x^4 - 1) dx ∫-∞^0 e^(2x) dx

Transcript text: \[ \int \frac{1}{x^{4}-1} d x \] \[ \int_{-\infty}^{0} e^{2 x} d x \]

Solution

Solution Steps

Solution Approach

For the first integral \(\int \frac{1}{x^{4}-1} dx\), we can use partial fraction decomposition to express the integrand as a sum of simpler fractions, which can then be integrated individually.
For the second integral \(\int_{-\infty}^{0} e^{2x} dx\), we need to evaluate the improper integral by finding the limit as the lower bound approaches \(-\infty\).

Step 1: 첫 번째 적분 계산

주어진 적분 \(\int \frac{1}{x^{4}-1} dx\)의 결과는 다음과 같습니다: \[ \int \frac{1}{x^{4}-1} dx = \frac{1}{4} \log(x - 1) - \frac{1}{4} \log(x + 1) - \frac{1}{2} \tan^{-1}(x) + C \] 여기서 \(C\)는 적분 상수입니다.

Step 2: 두 번째 적분 계산

두 번째 적분 \(\int_{-\infty}^{0} e^{2x} dx\)의 결과는 다음과 같습니다: \[ \int_{-\infty}^{0} e^{2x} dx = \frac{1}{2} \]

Final Answer

\[ \text{첫 번째 적분 결과: } \frac{1}{4} \log(x - 1) - \frac{1}{4} \log(x + 1) - \frac{1}{2} \tan^{-1}(x) + C \] \[ \text{두 번째 적분 결과: } \frac{1}{2} \] \(\boxed{\frac{1}{2}}\)

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