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  • Provide Solution For R.D.Sharma Maths Class 12 Chapter 18 Indefinite Integrals Exercise 18.25 Question 40 Maths Textbook Solution.

Provide Solution For R.D.Sharma Maths Class 12 Chapter 18  Indefinite Integrals Exercise 18.25 Question 40 Maths Textbook Solution.

Answers (1)

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

Given: \int \frac{x^{2}\tan ^{-1}x}{1+x^{2}}dx

Hint: Let 1+x^{2}=u and \tan ^{-1}x=t

Solution:

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

Let 1+x^{2}=u\Rightarrow 2xdx=dx

                \begin{aligned} &\int 1 \tan ^{-1} x=x \tan ^{-1} x-\int \frac{d u}{2 u} \\ &\Rightarrow x \tan ^{-1} x-\frac{1}{2} \ln u \\ &\Rightarrow x \tan ^{-1} x-\frac{1}{2} \ln \left(1+x^{2}\right) \end{aligned}

\int \frac{\tan ^{-1}x}{1+x^{2}}dx

Let,\tan ^{-1}x=\Rightarrow dt=\frac{1}{1+x^{2}}dx

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

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