Question

\[\int_0^1 \frac{1}{1 + 2x + 2 x^2 + 2 x^3 + x^4}dx\]

Answer 1

\[\int_0^1 \frac{1}{1 + 2x + 2 x^2 + 2 x^3 + x^4}dx\]
\[ = \int_0^1 \frac{1}{\left( x^2 + 1 \right)^2 + 2x\left( x^2 + 1 \right)}dx\]
\[ = \int_0^1 \frac{1}{\left( x^2 + 1 \right)\left( x^2 + 1 + 2x \right)}dx\]
\[ = \int_0^1 \frac{1}{\left( x^2 + 1 \right) \left( x + 1 \right)^2}dx\]

Let 

\[\frac{1}{\left( x + 1 \right)^2 \left( x^2 + 1 \right)} = \frac{A}{x + 1} + \frac{B}{\left( x + 1 \right)^2} + \frac{Cx + D}{x^2 + 1}\]
\[ \Rightarrow 1 = A\left( x + 1 \right)\left( x^2 + 1 \right) + B\left( x^2 + 1 \right) + \left( Cx + D \right) \left( x + 1 \right)^2\]

Putting x = −1, we have

1 = 2B

\[\Rightarrow B = \frac{1}{2}\]

Putting x = 0, we have

A + B + D = 1              .....(2)

Equating coefficient of x³ on both sides, we have

A + C = 0                    .....(3)

Equating coefficient of x² on both sides, we have

A + B + 2C + D = 0               .....(4)

⇒ 2C = −1               [Using (1)]

\[\Rightarrow C = - \frac{1}{2}\]

\[\therefore A = \frac{1}{2}\]

Putting

\[A = \frac{1}{2}, B = \frac{1}{2}\] and

\[C = - \frac{1}{2}\]  in (4), we have
D = 0

\[\therefore \int_0^1 \frac{1}{\left( x + 1 \right)^2 \left( x^2 + 1 \right)}dx\]
\[ = \int_0^1 \frac{\frac{1}{2}}{x + 1}dx + \int_0^1 \frac{\frac{1}{2}}{\left( x + 1 \right)^2}dx + \int_0^1 \frac{- \frac{1}{2}x}{x^2 + 1}\]
\[ = \left.\frac{1}{2} \log\left( x + 1 \right)\right|_0^1 + \left.\frac{1}{2} \times \left( - \frac{1}{x + 1} \right)\right|_0^1 - \frac{1}{4} \int_0^1 \frac{2x}{x^2 + 1}dx\]
\[ = \frac{1}{2}\left( \log2 - \log1 \right) - \frac{1}{2}\left( \frac{1}{2} - 1 \right) - \left.\frac{1}{4} \log\left( x^2 + 1 \right)\right|_0^1 \]
\[ = \frac{1}{2}\log2 + \frac{1}{4} - \frac{1}{4}\left( \log2 - \log1 \right) ................\left( \log1 = 0 \right)\]

\[= \frac{1}{2}\log 2 + \frac{1}{4}\log e - \frac{1}{4}\log2\]
\[ = \frac{1}{4}\log 2 + \frac{1}{4}\log e\]
\[ = \frac{1}{4}\left( \log 2 + \log e \right)\]
\[ = \frac{1}{4}\log\left( 2e \right)\]