Question

Solve : $\underset{0}{\stackrel{1}{\int }}{\cot }^{-1}(1+x+x^2)dx$

Answer 1

$\int {\cot }^{-1}(1+x+x^2)dx$

$=x{\cot }^{-1}(1+x+x^2)-\int \frac{(-2x-1)x}{(x^2+x+1{)}^2+1}dx$

$=x{\cot }^{-1}(1+x+x^2)-\int (-\frac{2x^2}{(x^2+x+1{)}^2+1}-\frac{x}{(x^2+x+1{)}^2+1})dx$

$=x{\cot }^{-1}(1+x+x^2)+2\int \frac{x^2}{(x^2+x+1{)}^2+1}dx+\int \frac{x}{(x^2+x+1{)}^2+1}dx$

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$\int \frac{x^2}{(x^2+x+1{)}^2+1}dx$

$=\frac{1}{5}\int \frac{2x-1}{x^2+1}dx-\frac{2}{5}\int \frac{x-1}{x^2+2x+2}dx$

$=-\frac{\ln (x^2+2x+2)}{5}+\frac{\ln (x^2+1)}{5}+\frac{4{\tan }^{-1}(x+1)}{5}-\frac{{\tan }^{-1}\left(x\right)}{5}$

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$\int \frac{x}{(x^2+x+1{)}^2+1}dx$

$=\frac{1}{5}\int \frac{x+2}{x^2+1}dx-\frac{1}{5}\int \frac{x+4}{x^2+2x+2}dx$

$=-\frac{\ln (x^2+2x+2)}{10}+\frac{\ln (x^2+1)}{10}-\frac{3{\tan }^{-1}(x+1)}{5}+\frac{2{\tan }^{-1}\left(x\right)}{5}$

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$=x{\cot }^{-1}(1+x+x^2)+2\int \frac{x^2}{(x^2+x+1{)}^2+1}dx+\int \frac{x}{(x^2+x+1{)}^2+1}dx$

$=x{\cot }^{-1}(1+x+x^2)-\frac{\ln (x^2+2x+2)}{2}+\frac{\ln (x^2+1)}{2}+{\tan }^{-1}(x+1)+C$