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Question Number 145941 by Mathspace last updated on 09/Jul/21

$$find{\int }_0^{\mathrm{\infty }}{e}^{-3x}log(1+x^3)dx$$

Answered by ArielVyny last updated on 24/Jul/21

$$\frac{3x^2}{1+x^3}=3\mathrm{\Sigma }{(-1)}^n{x}^{3n+2}$$$$ln(1+x^3)=\mathrm{\Sigma }3{(-1)}^n\frac{1}{3n+3}{x}^{3n+3}=\mathrm{\Sigma }\frac{{(-1)}^n}{n+1}{x}^{3n+3}$$$${\int }_0^{\mathrm{\infty }}{e}^{-3x}ln(1+x^3)=\mathrm{\Sigma }\frac{{(-1)}^n}{n+1}{\int }_0^{\mathrm{\infty }}{e}^{-3x}{x}^{3n+3}dx$$$$3x=t\to 3dx=dt$$$${\int }_0^{\mathrm{\infty }}{e}^{-t}{\left(\frac{t}{3}\right)}^{3n+3}\frac{1}{3}dt={\left(\frac{1}{3}\right)}^{3n+4}{\int }_0^{\mathrm{\infty }}{e}^{-t}{t}^{3n+3}dt$$$$={\left(\frac{1}{3}\right)}^{3n+4}\mathrm{\Gamma }(3n+4)={\left(\frac{1}{3}\right)}^{3n+4}(3n+3)!$$$${\int }_0^{\mathrm{\infty }}{e}^{-3x}ln(1+x^3)dx=\mathrm{\Sigma }\frac{{(-1)}^n}{n+1}{\left(\frac{1}{3}\right)}^{3n+4}(3n+3)!$$$$=\mathrm{\Sigma }\frac{{(-1)}^n(3n+3)}{n+1}{\left(\frac{1}{3}\right)}^{3n+4}(3n+2)!$$$$=\mathrm{\Sigma }{(-1)}^{n}3\times 3^{-(3n+4)}(3n+2)!$$$$=\mathrm{\Sigma }{(-1)}^n{\left(\frac{1}{3}\right)}^{3n+3}(3n+2)!$$$$=\frac{1}{27}\mathrm{\Sigma }{(-1)}^n{\left(\frac{1}{27}\right)}^n(3n+2)!$$

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