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Question Number 130835 by mathmax by abdo last updated on 29/Jan/21

$$\mathrm{calculate}{\int }_0^1\frac{\ln (1+{\mathrm{x}}^4)}{{\mathrm{x}}^2}\mathrm{dx}$$

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

$${\int }_0^1\frac{log(1+x^4)}{x^2}dx$$$$=\underset{n=1}{\sum ^{\mathrm{\infty }}}{(-1)}^{n+1}{\int }_0^1\frac{x^{4n-2}}{n}dx$$$$=\underset{n=1}{\sum ^{\mathrm{\infty }}}\frac{{(-1)}^{n+1}}{n(4n-1)}=\underset{n=1}{\sum ^{\mathrm{\infty }}}\frac{{(-1)}^{n+1}}{n}-4\underset{n=1}{\sum ^{\mathrm{\infty }}}\frac{{(-1)}^{n+1}}{4n-1}$$$$=log\left(2\right)-4(\frac{1}{3}-\frac{1}{7}+\frac{1}{11}-\frac{1}{15}+...)$$$$=log\left(2\right)-\frac{1}{2}(\underset{n=0}{\sum ^{\mathrm{\infty }}}\frac{1}{n+\frac{3}{8}}-\frac{1}{n+\frac{7}{8}})$$$$=log\left(2\right)-\frac{1}{2}\psi \left(\frac{7}{8}\right)+\frac{1}{2}\psi \left(\frac{3}{8}\right)$$

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