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Explain Solution R.D.Sharma Class 12 Chapter 18 Indefinite Integrals Exercise 18.3 Question 4 Maths Textbook Solution.

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Answer: \frac{-1}{2(x+1)^{2}}-\frac{2}{3(x+1)^{3}}+c

Hint: \text { To solve this equation we will spilt } x+3 \text { and then we use } \int \frac{1}{a x+b} d x \text { formula }

Given: \int \frac{x+3}{(x+1)^{4}} d x

Solution: \frac{x+3}{(x+1)^{4}}=\frac{x+1+2}{(x+1)^{4}}

\begin{aligned} &=\frac{x+1}{(x+1)^{4}}+\frac{2}{(x+1)^{4}}=\frac{1}{(x+1)^{3}}+\frac{2}{(x+1)^{4}} \\ &=\int \frac{x+3}{(x+1)^{4}} d x \end{aligned}

\begin{aligned} &=\int \frac{1}{(x+1)^{3}} d x+2 \int \frac{1}{(x+1)^{4}} d x\left[\int(a x+b)^{n}=\frac{1}{a(n+1)}(a x+b)^{n+1}+c, n \neq 1\right] \\ &=\frac{1}{1} \times \frac{1}{1-3}(x+1)^{1-3}+2 \frac{1}{1} \times \frac{1}{1-4}(x+1)^{1-4}+c \\ &=\frac{-1}{2}(x+1)^{-2}-\frac{2}{3}(x+1)^{-3}+c \\ &=\frac{-1}{2(x+1)^{2}}-\frac{2}{3(x+1)^{3}}+c \end{aligned}

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