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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.

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

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

So$A=-1,B=2$

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}

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