#### Provide Solution For R.D.Sharma Maths Class 12 Chapter 19 definite Integrals Exercise 19.1 Question 33 Maths Textbook Solution.

$\frac{3}{4}log3-log2$

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

Given:

$\int_{1}^{3} \frac{\log x}{(x+1)^{2}} d x=\int_{1}^{3} \log x \frac{1}{(x+1)^{2}} d x$

Integrating by parts then

\begin{aligned} &\int_{1}^{3} \log x \cdot \frac{1}{(x+1)^{2}} d x=\left[\log x \int \frac{1}{(x+1)^{2}} d x\right]_{1}^{3}-\int_{1}^{3}\left\{\frac{d}{d x}(\log x) \cdot \int \frac{1}{(x+1)^{2}} d x\right\} d x \\ &=\left[\log x \int(x+1)^{-2} d x\right]_{1}^{3}-\int_{1}^{3}\left\{\frac{1}{x} \cdot \int(x+1)^{-2} d x\right\} d x \\ &=\left[\log x\left(\frac{(x+1)^{-2+1}}{-2+1}\right)\right]_{1}^{3}-\int_{1}^{3} \frac{1}{x}\left(\frac{(x+1)^{-2+1}}{-2+1}\right) d x \end{aligned}

\begin{aligned} &=\left[\log x\left(\frac{(x+1)^{-1}}{-1}\right)\right]_{1}^{3}-\int_{1}^{3} \frac{1}{x}\left(\frac{(x+1)^{-1}}{-1}\right) d x \\ &=-\left[\frac{\log x}{x+1}\right]_{1}^{3}+\int_{1}^{3} \frac{1}{x(x+1)} d x \\ &=-\left[\frac{\log 3}{3+1}-\frac{\log 1}{1+1}\right]_{1}^{3}+\int_{1}^{3} \frac{1}{x(x+1)} d x \end{aligned}

\begin{aligned} &=-\left[\frac{\log 3}{4}-\frac{\log 1}{2}\right]+\int_{1}^{3} \frac{1}{x(x+1)} d x \\ &=-\left[\frac{\log 3}{4}-0\right]+\int_{1}^{3} \frac{1}{x(x+1)} d x \\ &=-\frac{1}{4} \log 3+\int_{1}^{3} \frac{1}{x(x+1)} d x \ldots . .(1) \\ &\therefore \int_{1}^{3} \frac{1}{x(x+1)} d x \end{aligned}

To solve this integral first we need to find its partial fraction then we will integrate and put the limits. So

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

Equating coefficient of x from both sides

$0=A+B\Rightarrow A=-B\Rightarrow B=-A$

Again Equating coefficient of constant term from both side, then

$1=A$

$\Rightarrow B=-A=-1$

$\therefore A=1,B=-1$

Thus

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

Now

$\int_{1}^{3}\left(\frac{1}{x}-\frac{1}{x+1}\right) d x=\int_{1}^{3} \frac{1}{x} d x-\int_{1}^{3} \frac{1}{x+1} d x$

put $x+1=u$

$\Rightarrow dx=du$ in the 2nd integral, then when x=1

$\Rightarrow u=2$ and when x=3,u=4 then

\begin{aligned} &\int_{1}^{3}\left(\frac{1}{x}-\frac{1}{x+1}\right) d x=\int_{1}^{3} \frac{1}{x} d x-\int_{2}^{4} \frac{1}{u} d u \\ &=[\log |x|]_{1}^{3}-[\log |u|]_{2}^{4} \\ &{\left[\because \int \frac{1}{x} d x=\log |x|\right]} \end{aligned}

\begin{aligned} &=[\log 3-\log 1]-[\log 4-\log 2] \\ &=[\log 3-0]-\left[\log \frac{4}{2}\right] \\ &=\log 3-\log 2 \ldots \ldots . .(2) \end{aligned}

putting the value of this integral eq(2) in eq(1) then

$\int_{1}^{3} \frac{\log x}{(x+1)^{2}} d x=-\frac{1}{4} \log 3+\log 3-\log 2$

\begin{aligned} &=\left(-\frac{1}{4}+1\right) \log 3-\log 2 \\ &=\left(\frac{-1+4}{4}\right) \log 3-\log 2 \\ &=\frac{3}{4} \log 3-\log 2 \end{aligned}