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Question Number 142318 by mnjuly1970 last updated on 29/May/21

$$Nice...\succcurlyeq \succcurlyeq \succcurlyeq \ast \ast \ast \preccurlyeq \preccurlyeq \preccurlyeq ...Calculus$$$$\mathrm{\Omega }:={\int }_0^1\frac{(1-\sqrt[3]{x})(1-\sqrt[5]{x})(1-\sqrt[7]{x})}{ln\left(\sqrt[3]{x}\right)}dx=?$$$$....m.n$$

Answered by Dwaipayan Shikari last updated on 29/May/21

$$\mathcal{K}\left(a\right)={\int }_0^1\frac{(1-x^a)(1-\sqrt[5]{x})(1-\sqrt[7]{x})}{log\left(x\right)}dx$$$${\mathcal{K}}^{\prime }\left(a\right)=-{\int }_0^1{x}^a(1-\sqrt[5]{x})(1-\sqrt[7]{x})dx=-{\int }_0^1{x}^a-x^{a+\frac{1}{5}}-x^{a+\frac{1}{7}}+x^{a+\frac{12}{35}}dx$$$${\mathcal{K}}^{\prime }\left(a\right)=-\frac{1}{a+1}+\frac{1}{a+\frac{6}{5}}+\frac{1}{a+\frac{8}{7}}-\frac{1}{a+\frac{47}{35}}$$$$\mathcal{K}\left(a\right)=log(\frac{a+\frac{6}{5}}{a+1}.\frac{a+\frac{8}{7}}{a+\frac{47}{35}})+C$$$$\mathcal{K}\left(0\right)=0\Rightarrow C=log\left(\frac{47}{48}\right)$$$$\mathcal{K}\left(a\right)=log(\frac{a+\frac{6}{5}}{a+1}.\frac{a+\frac{8}{7}}{a+\frac{47}{35}}.\frac{47}{48})$$

Commented by mnjuly1970 last updated on 29/May/21

$$thankyousomuch...$$

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