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Explain Solution R.D.Sharma Class 12 Chapter 18 Indefinite Integrals Exercise 18.25 Question 28 Maths Textbook Solution.

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Answer: -\frac{\log x}{(x+1)}+\log x-\log (x+1)

Hint:

Given: \int \frac{log\: x}{\left ( x+1 \right )^{2}}dx

Solution:

             I= \frac{log\: x}{\left ( x+1 \right )^{2}}dx

               =log\: x\int \frac{1}{\left ( x+1 \right )^{2}}dx-\int \left ( \frac{1}{x}\int \frac{1}{\left ( x+1 \right )^{2}} dx\right )dx

                =log\: x\left ( \frac{-1}{x+1} \right )+\int \frac{1}{x}\frac{1}{x+1}dx

                =-\frac{log\: x}{\left ( x+1 \right )}+I_{1}

                =I_{1}=\int \frac{1}{x\left ( x+1 \right )}dx

                  Using partial fractions

                Let \frac{1}{x\left ( x+1 \right )}=\frac{A}{x}+\frac{B}{x+1}

                            1=A\left ( x+1 \right )+B\left ( x \right )

                            1=Ax+A+Bx

                            1=\left ( A+B \right )x+A

                             \Rightarrow A+B=0

                             \Rightarrow A=1

                             \Rightarrow B=-1

              I_{1}=\int \left [ \frac{1}{x} -\frac{1}{x+1}\right ]dx

                =log\: x-log\left ( x+1 \right )

             I=\frac{log\, x}{x+1}+log\: x-log\left ( x+1 \right )

 

           

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