Question

${\int }_0^1\frac{1}{1+2x+2x^2+2x^3+x^4}dx$

Answer 1

$$\begin{gathered} {\int }_0^1\frac{1}{1+2x+2x^2+2x^3+x^4}dx \\ ={\int }_0^1\frac{1}{{\left(x^2+1\right)}^2+2x\left(x^2+1\right)}dx \\ ={\int }_0^1\frac{1}{\left(x^2+1\right)\left(x^2+1+2x\right)}dx \\ ={\int }_0^1\frac{1}{\left(x^2+1\right){\left(x+1\right)}^2}dx \end{gathered}$$

Let
$$\begin{gathered} \frac{1}{{\left(x+1\right)}^2\left(x^2+1\right)}=\frac{A}{x+1}+\frac{B}{{\left(x+1\right)}^2}+\frac{Cx+D}{x^2+1} \\ \Rightarrow 1=A\left(x+1\right)\left(x^2+1\right)+B\left(x^2+1\right)+\left(Cx+D\right){\left(x+1\right)}^2 \end{gathered}$$

Putting x = −1, we have

1 = 2B $\Rightarrow B=\frac{1}{2}$ .....(1)

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)]

$\Rightarrow C=-\frac{1}{2}$

$\therefore A=\frac{1}{2}$ [Using (3)]

Putting $A=\frac{1}{2},B=\frac{1}{2}$ and $C=-\frac{1}{2}$ in (4), we have

D = 0

$$\begin{gathered} \therefore {\int }_0^1\frac{1}{{\left(x+1\right)}^2\left(x^2+1\right)}dx \\ ={\int }_0^1\frac{{\displaystyle \frac{1}{2}}}{x+1}dx+{\int }_0^1\frac{{\displaystyle \frac{1}{2}}}{{\left(x+1\right)}^2}dx+{\int }_0^1\frac{-{\displaystyle \frac{1}{2}}x}{x^2+1} \\ =\frac{1}{2}{\overline{)\log \left(x+1\right)}}_0^1+\frac{1}{2}\times {\overline{)\left(-\frac{1}{x+1}\right)}}_0^1-\frac{1}{4}{\int }_0^1\frac{2x}{x^2+1}dx \\ =\frac{1}{2}\left(\log 2-\log 1\right)-\frac{1}{2}\left(\frac{1}{2}-1\right)-\frac{1}{4}{\overline{)\log \left(x^2+1\right)}}_0^1 \\ =\frac{1}{2}\log 2+\frac{1}{4}-\frac{1}{4}\left(\log 2-\log 1\right)\left(\log 1=0\right) \end{gathered}$$
$$\begin{gathered} =\frac{1}{2}\log 2+\frac{1}{4}\log e-\frac{1}{4}\log 2 \\ =\frac{1}{4}\log 2+\frac{1}{4}\log e \\ =\frac{1}{4}\left(\log 2+\log e\right) \\ =\frac{1}{4}\log \left(2e\right) \end{gathered}$$