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  • Need Solution for R.D.Sharma Maths Class 12 Chapter 19 definite Integrals Exercise 19.1 Question 55 Maths Textbook Solution.

Need Solution for R.D.Sharma Maths Class 12 Chapter 19 definite Integrals Exercise 19.1 Question 55 Maths Textbook Solution.

Answers (1)

Answer: log\left ( \frac{32}{27} \right )

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

Given:\int_{1}^{2} \frac{x}{(x+1)(x+2)} d x

Solution:

            \int_{1}^{2} \frac{x}{(x+1)(x+2)} d x                                                                    ..................(1)

To solve this integral, first we will final its partial fraction then integrate it with the given limits.

So,\frac{x}{\left ( x+1 \right )\left ( x+2 \right )}=\frac{A}{x+1}+\frac{B}{x+2}                                                        ............................(2)

\begin{aligned} &x=\frac{A(x+1)(x+2)}{(x+1)}+\frac{B(x+1)(x+2)}{(x+2)} \\ &\Rightarrow x=A(x+2)+B(x+1) \\ &\Rightarrow x=A x+2 A+B x+B \\ &\Rightarrow x=(A+B) x+2 A+B \end{aligned}

Equating coefficient of x from both sides,

1=A+B                                                                                                ........................(a)

And again equating coefficients of constant term from both sides, then

0=2A+B\Rightarrow 2A=-B\Rightarrow B=-2A                                            ........................(b)

Put the value of A from (b) in (a) then

\begin{gathered} 1=A+(-2 A)=A-2 A=-A \\ A=-1 \\ \Rightarrow B=-2 A=-2(-1)=2 \end{gathered}

SoA=-1,B=2

Then equation (2) becomes

\frac{x}{\left ( x+1 \right )\left ( x+2 \right )}=\frac{-1}{x+1}+\frac{2}{x+2}

Now equation (1)

\Rightarrow \int_{1}^{2} \frac{x}{(x+1)(x+2)} d x=\int_{1}^{2} \frac{-1}{x+1} d x+2 \int_{1}^{2} \frac{1}{x+2} d x                                            ..................(3)

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

Putting x+1=t\Rightarrow dx=dt

When x=1  then t=1+1=2

And When x=2 then t=2+1=3

Then \left.\left.\int_{1}^{2} \frac{1}{x+1} d x=\int_{1}^{2} \frac{1}{t} d t=[\log \mid t]\right]_{2}^{3} \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \quad\left[\int \frac{1}{x} d x=\log \mid x\right]\right]

=\left [ log\left | 3 \right |-log\left | 2 \right | \right ]

=\left [ log3-log2 \right ]                                                                            .......................(4)

and \int_{1}^{2}\frac{1}{x+1}dx

Putting x+2=u\Rightarrow dx=du

When x=1 then u=2+1=3

And x=2  then  u=2+2=4

Then\begin{aligned} \int_{1}^{2} \frac{1}{x+2} d x &=\int_{3}^{4} \frac{1}{u} d u=[\log |u|]_{3}^{4} \\ &=\log |4|-\log |3|=\log 4-\log 3 \end{aligned}                                        ......(5)

\begin{aligned} &\int_{1}^{2} \frac{x}{(x+1)(x+2)} d x=-\int_{1}^{2} \frac{1}{x+1} d x+2 \int_{1}^{2} \frac{1}{x+2} d x \\ &=-[\log 3-\log 2]+2[\log 4-\log 3] \\ &=-\log 3+\log 2+2 \log 2^{2}-2 \log 3 \; \; \; \; \; \; \; \; \; \; \; \; \quad\left[\log a^{m}=m \log a\right] \\ &=-3 \log 3+\log 2+4 \log 2 \end{aligned}

\begin{aligned} &=-3 \log 3+5 \log 2\\ &\begin{aligned} &=5 \log 2-3 \log 3 \\ &=\log 2^{5}-\log 3^{3} \end{aligned}\; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \left[\log a-\log b=\log \frac{a}{b}\right]\\ &=\log 32-\log 27 \end{aligned}

=log\left ( \frac{32}{27} \right )

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