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Question 129
∫24e2xex−1dx=23(e4−1)32+2(e4−1)12−23(e2−1)32−2(e2−1)12\int_{2}^{4} \frac{e^{2 x}}{\sqrt{e^{x}-1}} d x=\frac{2}{3}\left(e^{4}-1\right)^{\frac{3}{2}}+2\left(e^{4}-1\right)^{\frac{1}{2}}-\frac{2}{3}\left(e^{2}-1\right)^{\frac{3}{2}}-2\left(e^{2}-1\right)^{\frac{1}{2}}∫24ex−1e2xdx=32(e4−1)23+2(e4−1)21−32(e2−1)23−2(e2−1)21 .
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