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Provide Solution For R.D.Sharma Maths Class 12 Chapter 18  Indefinite Integrals Exercise 18.25 Question 48 Maths Textbook Solution.

Answers (1)

Answer: \tan ^{-1}\sqrt{x}\left ( 1+x \right )-\sqrt{x}+c

Given: \int \left ( \tan ^{-1}\sqrt{x} \right )dx

Hint:Put\sqrt{x}=t

Solution:

            I= \left ( \tan ^{-1}\sqrt{x} \right )dx

Put \sqrt{x}=t\Rightarrow dx=2tdt

         \begin{aligned} &I=\int \tan ^{-1} t 2 t d t \\ &I=\tan ^{-1} t 2 \cdot \frac{t^{2}}{2}-\int \frac{t^{2}+1-1}{1+t^{2}} d t \\ &=t^{2} \tan ^{-1} t-\int d t+\int \frac{1}{1+t^{2}} d t \\ &=t^{2} \tan ^{-1} t-t+\tan ^{-1} t+c \\ &=\tan ^{-1} t\left(1+t^{2}\right)-t+c \\ &=\tan ^{-1} \sqrt{x}(1+x)-\sqrt{x}+c \end{aligned}

          

 

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