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Question Number 57229 by maxmathsup by imad last updated on 31/Mar/19

$$give{\int }_0^1\frac{x^5}{x^3+1}dxatformofserie$$

Commented by maxmathsup by imad last updated on 01/Apr/19

$$letI={\int }_0^1\frac{x^5}{x^3+1}dx\Rightarrow I={\int }_0^1{x}^5\left(\sum _{n=0}^{\mathrm{\infty }}{(-x)}^{3n}\right)dx=\sum _{n=0}^{\mathrm{\infty }}{(-1)}^n{\int }_0^1{x}^{3n+5}dx$$$$=\sum _{n=0}^{\mathrm{\infty }}{(-1)}^n\frac{1}{3n+6}=\frac{1}{3}\sum _{n=0}^{\mathrm{\infty }}\frac{{(-1)}^n}{n+2}$$$$letfindthevalueofIwehave\sum _{n=0}^{\mathrm{\infty }}\frac{{(-1)}^n}{n+2}{=}_{n+2=p}\sum _{p=2}^{\mathrm{\infty }}\frac{{(-1)}^{p-2}}{p}$$$$=\sum _{p=2}^{\mathrm{\infty }}\frac{{(-1)}^p}{p}=\sum _{p=1}^{\mathrm{\infty }}\frac{{(-1)}^p}{p}+1=-ln\left(2\right)+1\Rightarrow I=\frac{1}{3}(1-ln\left(2\right))$$

Answered by tanmay.chaudhury50@gmail.com last updated on 01/Apr/19

$$t=1+x^3$$$$dt=3x^{2}dx\to \frac{dt}{3}=x^{2}dx$$$${\int }_0^1\frac{x^3\times x^{2}dx}{1+x^3}$$$${\int }_1^2\frac{t-1}{t}\times \frac{dt}{3}$$$$\frac{1}{3}{\int }_1^2(1-\frac{1}{t})dt$$$$\frac{1}{3}\mid t-lnt{\mid }_1^2$$$$\frac{1}{3}\{(2-1)-(ln2-ln1)\}$$$$\frac{1}{3}\{1-ln2)$$

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