The integral _0^1 ^-1(1+x+x^2)dx is equal to :

Question

The integral $\int _{0}^{1}{\cot }^{-1}(1+x+{x}^{2})dx$ is equal to :

Accepted Answer

Let

$I={\int }_{1}^{0}{\cot }^{-1}(1+x+{x}^{2})\, dx$

We will use the standard identity

${\cot }^{-1}t={	an }^{-1}(\frac{1}{t})\, t>0$

Since $1+x+{x}^{2}>0$ on $[0,1]$, we can write

$I={\int }_{1}^{0}{	an }^{-1}(\frac{1}{1+x+{x}^{2}})\, dx$

Now use the formula

${	an }^{-1}a-{	an }^{-1}b={	an }^{-1}(\frac{a-b}{1+ab})$

We try to express

$\frac{1}{1+x+{x}^{2}}$

in a suitable form. Observe that

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

Hence,

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

So,

$I={\int }_{1}^{0}[{	an }^{-1}(x+1)-{	an }^{-1}x]dx$

Therefore,

$I={\int }_{1}^{0}{	an }^{-1}(x+1)\, dx-{\int }_{1}^{0}{	an }^{-1}x\, dx$

In the first integral, put $u=x+1$. Then when $x=0$, $u=1$, and when $x=1$, $u=2$. So

${\int }_{1}^{0}{	an }^{-1}(x+1)\, dx={\int }_{2}^{1}{	an }^{-1}u\, du$

Thus,

$I={\int }_{2}^{1}{	an }^{-1}u\, du-{\int }_{1}^{0}{	an }^{-1}x\, dx={\int }_{2}^{1}{	an }^{-1}x\, dx-{\int }_{1}^{0}{	an }^{-1}x\, dx$

Now use

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

So,

${\int }_{2}^{1}{	an }^{-1}x\, dx={[x{	an }^{-1}x-\frac{1}{2}\log (1+{x}^{2})]}_{2}^{1}$

and

${\int }_{1}^{0}{	an }^{-1}x\, dx={[x{	an }^{-1}x-\frac{1}{2}\log (1+{x}^{2})]}_{1}^{0}$

Hence,

$I={[x{	an }^{-1}x-\frac{1}{2}\log (1+{x}^{2})]}_{2}^{1}-{[x{	an }^{-1}x-\frac{1}{2}\log (1+{x}^{2})]}_{1}^{0}$

This becomes

$I=(2{	an }^{-1}2-\frac{1}{2}\log 5)-({	an }^{-1}1-\frac{1}{2}\log 2)-({	an }^{-1}1-\frac{1}{2}\log 2)$

since at $x=0$, the value is $0$.

Now ${	an }^{-1}1=\frac{\pi }{4}$, so

$I=2{	an }^{-1}2-\frac{1}{2}\log 5-2(\frac{\pi }{4}-\frac{1}{2}\log 2)$

$I=2{	an }^{-1}2-\frac{1}{2}\log 5-\frac{\pi }{2}+\log 2$

Now combine the logarithms:

$\log 2-\frac{1}{2}\log 5=\frac{1}{2}\log 4-\frac{1}{2}\log 5=\frac{1}{2}\log (\frac{4}{5})=-\frac{1}{2}\log (\frac{5}{4})$

Therefore,

$I=2{	an }^{-1}2-\frac{1}{2}\log (\frac{5}{4})-\frac{\pi }{2}$

So the correct answer is

$2{	an }^{-1}2-\frac{1}{2}{\log }_{e}(\frac{5}{4})-\frac{\pi }{2}$

Hence, the correct option is:

Solutions · 1