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Please Solve R.D.Sharma Class 12 Chapter 18 Indefinite Integrals Exercise  18.25 Question 23 Maths Textbook Solution.

Answers (1)

Answer:

          =\frac{-[\log (x+2)+1]}{x+2}+c

Hint: Let

           x+2=t

Given: \int \frac{\log (x+2)}{(x+2)^{2}} d x

Solution:

           Let x+2=t \Rightarrow d x=d t

                 \begin{aligned} &\int \frac{\log (x+2)}{(x+2)^{2}} d x \Rightarrow \int \frac{\log t}{t^{2}} d t \Rightarrow \int \log t \frac{1}{t^{2}} d t \\ &=\log t \int \frac{1}{t^{2}} d t-\int \frac{d}{d t} \log t\left(\int \frac{1}{t^{2}} d t\right) d t+c \\ &=\log t\left(\frac{-1}{t}\right)-\int \frac{1}{t}\left(\frac{-1}{t}\right) d t+c \\ &=-\frac{\log t}{t}+\int \frac{1}{t^{2}} d t+c \\ &=-\frac{\log t}{t}+\left(\frac{-1}{t}\right)+c \\ &=\frac{-(\log t+1)}{t}+c \\ &=\frac{-[\log (x+2)+1]}{x+2}+c \end{aligned}

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