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Please Solve RD Sharma Class 12 Chapter 18 Indefinite Integrals Exercise18. 5 Question 1 Maths Textbook Solution.

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Answer:  \frac{1}{6}(2 x+3)^{\frac{3}{2}}-\frac{1}{2}(2 x+3)^{\frac{1}{2}}+C

Hint: Use Integration by partial fraction

Given: \int \frac{x+1}{2x+3}dx


= \int \frac{x+1}{2x+3}dx

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


=\frac{1}{2} \times \frac{1}{2} \int \frac{(2 x+3)^{\frac{1}{2}+1}}{\frac{1}{2}+1}-\frac{1}{2} \times \frac{1}{2} \int \frac{(2 x+3)^{\frac{-1}{2}+1}}{\frac{-1}{2}+1}                               \left[\therefore \int(a x+b)^{n} d x=\frac{\frac{1}{a}(a x+b)^{n+1}}{n+1}\right]

\begin{aligned} &=\frac{1}{2} \int \frac{(2 x+3)^{\frac{3}{2}}}{3}-\frac{1}{2} \times(2 x+3)^{\frac{1}{2}}+C \\ &=\frac{1}{6}(2 x+3)^{\frac{3}{2}}-\frac{1}{2}(2 x+3)^{\frac{1}{2}}+C \end{aligned}

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