Question

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

Answer 1

$\underset{0}{\overset{1}{\int }}\frac{dx}{1+2x+2x^2+2x^3+x^4}$

$=\underset{0}{\overset{1}{\int }}\frac{dx}{(1+x^2)(1+2x+x^2)}$

$=\underset{0}{\overset{1}{\int }}\frac{dx}{(1+x^2)(1+x{)}^2}$

$=\underset{0}{\overset{1}{\int }}\frac{Axdx}{(1+x^2)}$ $+\underset{0}{\overset{1}{\int }}\frac{Bdx}{(1+x)}$ $+\underset{0}{\overset{1}{\int }}\frac{Cdx}{(1+x{)}^2}$ {Let]........(1).

Then comparing both sides we get,

$1=Ax(1+x{)}^2+B(1+x^2\left)\right(1+x)+C(1+x^2)$.......(2).

Putting $x=-1$ in (2) we get,

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

Putting $x=i$ in (2) we get [where $i=\sqrt{-1}$]

$1=i.\left(2i\right)A$ [Since $(1+i{)}^2=2i$]

or, $A=-\frac{1}{2}$

Putting $x=0$ in (2) we get,

$1=B+C$

or, $B=\frac{1}{2}$ [Using value of $C$]

Now, from (1) using values of $A$, $B$ and $C$ we get,

$\underset{0}{\overset{1}{\int }}\frac{dx}{1+2x+2x^2+2x^3+x^4}$

$=\underset{0}{\overset{1}{\int }}-\frac{1}{2}\frac{xdx}{(1+x^2)}$ $+\underset{0}{\overset{1}{\int }}\frac{\frac{1}{2}dx}{(1+x)}$ $+\underset{0}{\overset{1}{\int }}\frac{\frac{1}{2}dx}{(1+x{)}^2}$

$=-\frac{1}{4}$ ${\left[\log \right(1+x^2\left)\right]}_0^1+\frac{1}{2}$ ${\left[\log \right(1+x\left)\right]}_0^1-\frac{1}{2}{\left[\frac{1}{(1+x)}\right]}_0^1$

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

$=\frac{1}{4}$$\log \left(2\right)+$ $\frac{1}{4}$.