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Need Solution for R.D.Sharma Maths Class 12 Chapter 18 Indefinite Integrals Exercise 18.3 Question 1 Maths Textbook Solution.

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Answer:\frac{(2 x-3)^{\6}}{12}+\frac{2}{9}(3 x+2)^{\frac{3}{2}}+c

Hint: To solve this equation we use \int(a x+b)^{n} formula

Given: \int(2 x-3)^{5}+\sqrt{3 x+2} d x

Solution:\int(a x+b)^{n} d x=\frac{1}{a(n+1)}(a x+b)^{n+1}+c ; n \neq-1

\begin{aligned} &=\int\left[(2 x-3)^{5}+(3 x+2)^{\frac{1}{2}}\right] d x \\ &=\int(2 x-3)^{5} d x+\int(3 x+2)^{\frac{1}{2}} d x \end{aligned}

\begin{aligned} &=\frac{1}{2(5+1)}(2 x-3)^{5+1}+\frac{1}{3\left(1+\frac{1}{2}\right)}(3 x+2)^{\frac{1}{2}+1}+c \\ &=\frac{1}{12}(2 x-3)^{6}+\frac{2}{9}(3 x+2)^{\frac{3}{2}}+c \end{aligned}

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