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Provide Solution For R.D.Sharma Maths Class 12 Chapter 18  Indefinite Integrals Exercise 18.25 Question 2 Maths Textbook Solution.

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Answer: x \log (x+1)-x+\log (x+1)+c

Hint:   We use integration by parts,

           \int u v d x=u \int v d x-\int \frac{d}{d x} u \int v d x

Given: Let I=\int \log (x+1) d x

Solution: I=\int 1 \times \log (x+1) d x

              I=\log (x+1) \int 1 d x-\int \frac{d}{d x} \log (x+1) \int 1 d x

We know that, \int 1 d x=x \& \frac{d}{d x} \log x=\frac{1}{x}

           =\log (x+1) \times x-\int \frac{1}{x+1} \times x

           \begin{aligned} &=\frac{x}{x+1}=1-\frac{1}{x+1} \\ &=x \log (x+1)-\int\left(1-\frac{1}{x+1}\right) d x \\ &=x \log (x+1)-x+\log (x+1)+c \end{aligned}


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