0-1-cot-1-1-x-x-2-dx-

Question Number 17203 by Arnab Maiti last updated on 02/Jul/17

\int_{0}^{1} \cot^{- 1} (1 - x + x^{2}) dx

Commented by prakash jain last updated on 02/Jul/17

take 1 as second function

\int \cot^{- 1} (1 - x + x^{2}) \cdot 1 d x

x \cot^{- 1} (1 - x + x^{2}) + \int \frac{x (2 x - 1)}{{(1 - x + x^{2})}^{2} + 1} d x

\int \frac{x (2 x - 1)}{{(1 - x + x^{2})}^{2} + 1} d x

1 + x^{2} + x^{4} - 2 x + 2 x^{2} - 2 x^{3} + 1

= x^{4} - 2 x^{3} + 3 x^{2} - 2 x + 2

= x^{4} + x^{2} - 2 x^{3} - 2 x + 2 x^{2} + 2

= x^{2} (x^{2} + 1) - 2 x (x^{2} + 1) + 2 (x^{2} + 1)

= (x^{2} + 1) (x^{2} - 2 x + 2)

\frac{2 x^{2} - x}{x^{2} + 1) (x^{2} - 2 x + 2)} = \frac{A x + B}{x^{2} + 1} + \frac{C x + D}{x^{2} - 2 x + 2}

x^{3} : A + C = 0 \Rightarrow A = - C

x^{2} : B - 2 A + D = 2

x : 2 A - 2 B + C = - 1

1 : 2 B + D = 0 \Rightarrow D = - 2 B

x^{2} : B - 2 A + D = 2

\Rightarrow B - 2 A - 2 B = 2 \Rightarrow B = - 2 A - 2

x : 2 A - 2 B + C = 1

2 A + 4 A + 4 - A = 1 \Rightarrow A = - 1

A = - 1, B = 0, C = 1, D = 0

\frac{x}{x^{2} + 1) (x^{2} - 2 x + 2)} = \frac{- x}{x^{2} + 1} + \frac{x}{x^{2} - 2 x + 2}

= (\frac{- x}{x^{2} + 1} + \frac{x}{x^{2} - 2 x + 2})

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

= - \frac{1}{2} \ln (x^{2} + 1)

\int \frac{x}{x^{2} - 2 x + 2} d x

= \frac{1}{2} \int \frac{2 x - 2 + 2}{x^{2} - 2 x + 2} d x

= \frac{1}{2} \int \frac{2 x - 2}{x^{2} - 2 x + 2} d x + \int \frac{1}{x^{2} - 2 x + 1 + 1} d x

\frac{1}{2} \int \frac{2 x - 2}{x^{2} - 2 x + 2} d x = \frac{1}{2} \ln (x^{2} - 2 x + 2)

\int \frac{1}{x^{2} - 2 x + 1 + 1} d x

\int \frac{1}{{(x - 1)}^{2} + 1} d x = \tan^{- 1} (x - 1)

\int \cot^{- 1} (1 - x + x^{2}) d x =

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

\frac{- 1}{2} \ln (x^{2} + 1) +

\frac{1}{2} \ln (x^{2} - 2 x + 2) + \tan^{- 1} (x - 1)]

Commented by prakash jain last updated on 02/Jul/17

x = 1

\cot^{- 1} 1 - \frac{1}{2} \ln 2 = \frac{π}{4} - \frac{1}{2} \ln 2

x = 0

\frac{1}{2} \ln 2 + \tan^{- 1} - 1 = \frac{1}{2} \ln 2 - \frac{π}{4}

\int_{0}^{1} \cot^{- 1} (1 - x + x^{2}) = \frac{π}{2} - \ln 2

Commented by Arnab Maiti last updated on 02/Jul/17

Ans = \frac{Π}{2} - \ln 2