# Evaluate the following definite integrals. int_{1/8}^1frac{dx}{xsqrt{1+x^{2/3}}}

Question
Integrals
Evaluate the following definite integrals.
$$\int_{1/8}^1\frac{dx}{x\sqrt{1+x^{2/3}}}$$

2020-12-15
$$\text{Consider the integral:}\\ \(\int_{1/8}^1\frac{dx}{x\sqrt{1+x^{2/3}}}\\ \(\text{Apply u-substitution: }u=\sqrt{1+x^{2/3}}\\ \(\int_{1/8}^1\frac{dx}{x\sqrt{1+x^{2/3}}}=\int_{\frac{5^{1/2}}{2}}^{\sqrt{2}}\frac{3}{u^2-1}du\\ \(=3\cdot\int_{\frac{5^{1/2}}{2}}^{\sqrt{2}}\frac{1}{u^2-1}du\\ \(=3\cdot\int_{\frac{5^{1/2}}{2}}^{\sqrt{2}}\frac{1}{-(-u^2+1)}du\\ \(=3(-\int_{\frac{5^{1/2}}{2}}^{\sqrt{2}}\frac{1}{-u^2+1}du\\ \(=3(-[\frac{\ln|u+1|}{2}-\frac{\ln|u-1|}{2}]_{\frac{5^{1/2}}{2}}^{\sqrt{2}})\\ \(=-3[\frac{1}{2}(\ln|u+1|-\ln|u-1|)]_{\frac{5^{1/2}}{2}}^{\sqrt{2}}\\ \(=-3\cdot\frac{\ln(\sqrt{2}+1)-\ln(\sqrt{2}-1)-\ln(\frac{\sqrt5}{2}+1)+\ln(\frac{\sqrt5}{2}-1)}{2}\\ \(\text{Answer in terms of logarithms:}\\ \(\int_{1/8}^1\frac{dx}{x\sqrt{1+x^{2/3}}}=-3\cdot\frac{\ln(\sqrt{2}+1)-\ln(\sqrt{2}-1)-\ln(\frac{\sqrt5}{2}+1)+\ln(\frac{\sqrt5}{2}-1)}{2}$$

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