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

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

Answers (1)

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}

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