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  • Please Solve R.D.Sharma Class 12 Chapter 18 Indefinite Integrals Exercise 18.25 Question 47 Maths Textbook Solution.

Please Solve R.D.Sharma Class 12 Chapter 18 Indefinite Integrals Exercise 18.25 Question 47 Maths Textbook Solution.

Answers (1)

Answer: \frac{1}{\sqrt{1+x^{2}}}\left ( x-\tan ^{-1}x \right )+c

Hint:  x=\tan \theta&\int uvdx=u\int vdx-\int \left ( \frac{d}{dx}u\int vdx \right )dx

Given: \int \frac{x\tan ^{-1}x}{\left ( 1+x^{2} \right )^{\frac{3}{2}}}dx

Solution:

           I= \frac{x\tan ^{-1}x}{\left ( 1+x^{2} \right )^{\frac{3}{2}}}dx

Put x=\tan \Theta \Rightarrow dx=\sec ^{2}\Theta \, d\: \Theta

\begin{gathered} I=\int \frac{\tan \theta \tan ^{-1}(\tan \theta) \sec ^{2} \theta d \theta}{\left(\sqrt{1+\tan ^{2} \theta}\right)^{\frac{3}{2}}} \\ =\int \frac{\tan \sec ^{2} \theta}{\sqrt{\sec ^{2} \theta} \sec ^{3} \theta} \\ =\int\left(\frac{\theta \tan \theta}{\sec \theta}\right) d \theta \end{gathered}

               \begin{aligned} &I=\int \theta \frac{\sin \theta}{\cos \theta} \times \cos \theta d \theta \\ &I=\theta \int \sin \theta d \theta-\int\left[\frac{d}{d \theta} \theta \int \sin \theta d \theta\right] d \theta \\ &=-\theta \cos \theta+\int(\cos \theta) d \theta \\ &=-\theta \cos \theta+\sin \theta+c \\ &=-\tan ^{-1} x\left(\frac{1}{\sqrt{1+x^{2}}}\right)+\frac{x}{\sqrt{1+x^{2}}}+c \\ &=\frac{1}{\sqrt{1+x^{2}}}\left(x-\tan ^{-1} x\right)+c \end{aligned}

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