#### Need Solution for R.D.Sharma Maths Class 12 Chapter 19 definite Integrals Exercise 19.1 Question 68 Maths Textbook Solution.

Answer: $\frac{1}{4}log\left ( 2e \right )$

Hint: Use indefinite formula then put the limit to solve this integral

Given: $\int_{0}^{1} \frac{1}{1+2 x+2 x^{2}+2 x^{3}+x^{4}} d x$

Solution: $\int_{0}^{1} \frac{1}{1+2 x+2 x^{2}+2 x^{3}+x^{4}} d x$

\begin{aligned} &=\int_{0}^{1} \frac{1}{x^{4}+2 x^{3}+2 x^{2}+2 x+1} d x \\ &=\int_{0}^{1} \frac{1}{x^{4}+x^{2}+x^{2}+2 x^{3}+2 x+1} d x \\ &=\int_{0}^{1} \frac{1}{x^{4}+x^{2}+2 x^{3}+x^{2}+2 x+1} d x \\ &=\int_{0}^{1} \frac{1}{x^{2}\left(x^{2}+1\right)+2 x\left(x^{2}+1\right)+\left(x^{2}+1\right)} d x \\ &=\int_{0}^{1} \frac{1}{\left(x^{2}+1\right)\left(x^{2}+2 x+1\right)} d x \\ &=\int_{0}^{1} \frac{1}{\left(x^{2}+1\right)(x+1)^{2}} d x \end{aligned}

$\left [ \left ( a+b \right )^{2}=a^{2}+2ab+b^{2} \right ]$

To solve this integral, first we have to find its partial fractions then integrate it by using indefinite integral formula then put the limits to get required answer.

So,

\begin{aligned} &\frac{1}{\left(x^{2}+1\right)(x+1)^{2}}=\frac{A}{x+1}+\frac{B}{(x+1)^{2}}+\frac{C x+D}{x^{2}+1} \\ &\Rightarrow 1=\frac{A(x+1)^{2}\left(x^{2}+1\right)}{x+1}+\frac{B(x+1)^{2}\left(x^{2}+1\right)}{(x+1)^{2}}+\frac{(C x+D)(x+1)^{2}\left(x^{2}+1\right)}{\left(x^{2}+1\right)} \\ &\Rightarrow 1=A(x+1)^{2}\left(x^{2}+1\right)+B\left(x^{2}+1\right)+(C x+D)(x+1)^{2} \\ &\Rightarrow 1=A\left(x^{3}+x+x^{2}+1\right)+B\left(x^{2}+1\right)+(C x+D)\left(x^{2}+2 x+1\right) \end{aligned}

$\left [ \left ( a+b \right )^{2}=a^{2}+b^{2}+2ab \right ]$

$\Rightarrow 1=A x^{3}+A x+A x^{2}+A+B x^{2}+B+C x^{3}+2 C x^{2}+C x+D x^{2}+2 D x+D$

Equating coefficient of  and constant term respectively

$0=A+C$

$0=A+B+2C+D$

$0=A+C+2D$

$1=A+B+D$

From (a) and (c)

$A+C+2D=0$

$\Rightarrow 0+2D=0$

$\Rightarrow D=0$

Put the value of D in (b) and (d)

$A+B+2C=0$

$A+B=1$

Subtracting (e) – (f) then

$A+B+2C=0$

$\frac{A+B=1}{2C=-1}$

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

Then $\left ( a \right )=>A=-C=-\left ( -\frac{1}{2} \right )=\frac{1}{2}$

And (d)

\begin{aligned} &\Rightarrow A+B+D=1 \\ &\Rightarrow \frac{1}{2}+B+0=1 \\ &\Rightarrow B=1-\frac{1}{2}=\frac{1}{2} \\ &A=\frac{1}{2}, B=\frac{1}{2}, C=-\frac{1}{2} \end{aligned}

\begin{aligned} &\frac{1}{\left(x^{2}+1\right)(x+1)^{2}}=\frac{1}{2(x+1)}+\frac{1}{2(x+1)^{2}}+\frac{\frac{-1}{2} x+0}{x^{2}+1} \\ &=\frac{1}{2(x+1)}+\frac{1}{2(x+1)^{2}}-\frac{1}{2} \frac{x}{x^{2}+1} \end{aligned}

Now $\int_{0}^{1} \frac{1}{1+2 x+2 x^{2}+2 x^{2}+x} d x=\int_{0}^{1}\left(\frac{1}{2(x+1)}+\frac{1}{2(x+1)^{2}}-\frac{1}{2} \frac{x}{x^{2}+1}\right) d x$

$=\frac{1}{2} \int_{0}^{1} \frac{1}{(x+1)} d x+\frac{1}{2} \int_{0}^{1} \frac{1}{(x+1)^{2}} d x-\frac{1}{2} \int_{0}^{1} \frac{x}{x^{2}} d x$            ..........(1)

Put $x+1=t\Rightarrow dx=dt$

When $x=0$ then $t=1$

And When $x=1$ then $t=2$

Then $\frac{1}{2} \int_{0}^{1} \frac{1}{(x+1)} d x+\frac{1}{2} \int_{0}^{1} \frac{1}{(x+1)^{2}} d x$

$=\frac{1}{2} \int_{1}^{2} \frac{1}{t} d t+\frac{1}{2} \int_{1}^{2} \frac{1}{t^{2}} d t=\frac{1}{2} \int_{1}^{2} \frac{1}{t} d t+\frac{1}{2} \int_{1}^{2} t^{-2} d t \quad\left[\int x^{n} d x=\frac{x^{n+1}}{n+1}, \int \frac{1}{x} d x=\log |x|\right]$

\begin{aligned} &=\frac{1}{2}[\log 2-\log 1]+\frac{1}{2}\left[\frac{t^{-1}}{-1}\right]_{1}^{2} \\ &=\frac{1}{2}[\log 2-\log 1]-\frac{1}{2}\left[\frac{1}{t}\right]_{1}^{2} \\ &=\frac{1}{2} \log 2-\frac{1}{2}\left[\frac{1}{2}-\frac{1}{1}\right] \\ &=\frac{1}{2} \log 2-\frac{1}{2}\left[\frac{1-2}{2}\right] \\ &=\frac{1}{2} \log 2-\frac{1}{2}\left[\frac{-1}{2}\right] \\ &=\frac{1}{2} \log 2+\frac{1}{4} \end{aligned}

And

$\frac{1}{2}\int_{0}^{1}\frac{x}{x^{2}+1}dx$

When $x=0$ then $p=1$ and

When $x=1$ then $p=2$

\begin{aligned} &\frac{1}{2} \int_{0}^{1} \frac{x}{x^{2}+1} d x=\frac{1}{2} \int_{1}^{2} \frac{1}{p} \frac{d p}{2} \\ &=\frac{1}{4} \int_{1}^{2} \frac{1}{p} d p \\ &=\frac{1}{4}[\log |p|]_{1}^{2} \\ &=\frac{1}{4}[\log 2-\log 1] \\ &=\frac{1}{4}[\log 2] \end{aligned} \quad\left[\begin{array}{l} \left.\frac{1}{x} d x=\log |x|\right] \\ \end{array}\right.$\left [ log1=0 \right ]$

Then From (1)

\begin{aligned} &\int_{0}^{1} \frac{1}{1+2 x+2 x^{2}+2 x^{3}+x^{4}} d x=\frac{1}{2} \log 2+\frac{1}{4}-\frac{1}{4} \log 2 \\ &=\log 2\left(\frac{1}{2}-\frac{1}{4}\right)+\frac{1}{4} \\ &=\log 2\left(\frac{2-1}{4}\right)+\frac{1}{4} \\ &=\frac{1}{4} \log 2+\frac{1}{4} \\ &=\frac{1}{4} \log 2+\frac{1}{4} \text { loge }[\because \text { loge }=1] \\ &=\frac{1}{4} l(\operatorname{og} 2+\log e) \\ &=\frac{1}{4} \log (2 e) \quad[\because \log m+\log n=\log (m n)] \end{aligned}