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

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

Answers (1)

Answer:

            \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}

 

 

Posted by

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