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  • Please Solve R.D.Sharma Class 12 Chapter 18 Indefinite Integrals Exercise Revision Exercise Question 34 Maths Textbbok Solution.

Please Solve R.D.Sharma Class 12 Chapter 18 Indefinite Integrals Exercise  Revision Exercise Question 34 Maths Textbbok Solution.

Answers (1)

Answer:

I=\frac{(2 x+3)^{2}}{8}-\frac{3(2 x+3)^{1}}{4}+c

Given:

\int x \sqrt{2 x+3} d x

Hint

 To solve the statement we have to put root statement to \mathrm{t}: \sqrt{2 x+3}=t

Solution: 

   \sqrt{2 x+3}=t

    put 2 x+3=t^{2}=>x=\frac{t^{2}-3}{2}

   2 d x=2 t d t

   d x=t d t

   I=\int \frac{t^{2}-3}{2} t d t

   I=\frac{1}{2} \int\left(t^{2}-3\right) t d t

   I=\frac{1}{2} \int\left(t^{3}-3 t\right) d t  

   I=\frac{1}{2}\left(\frac{t^{4}}{4}-\frac{3 t^{2}}{2}\right)

   I=\frac{t^{4}}{8}-\frac{3 t^{2}}{4}+c

   I=\frac{(2 x+3)^{2}}{8}-\frac{3(2 x+3)^{4}}{4}+c

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