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  • Explain Solution R.D.Sharma Class 12 Chapter 18 Indefinite Integrals Exercise Multiple Choice Questions Question 40 Maths Textbook Solution.

Explain Solution R.D.Sharma Class 12 Chapter 18 Indefinite Integrals Exercise Multiple Choice Questions Question 40 Maths Textbook Solution.

Answers (1)

Answer:

(x+1) \tan ^{-1} \sqrt{x}-\sqrt{x}+C

Given:

\int \tan ^{-1} \sqrt{x} d x

Hint:

Using integration by parts and \int \frac{1}{1+x^{2}} d x.

Explanation:

Let I=\int \tan ^{-1} \sqrt{x} d x

\Rightarrow I=\int \tan ^{-1} t .2 t d t                                                \left[\text { Put } \sqrt{x}=t \Rightarrow \frac{1}{2 \sqrt{x}} d x=d t \Rightarrow d x=2 \sqrt{x} d t \Rightarrow d x=2 t d t\right]

=2 \int \tan ^{-1} t \cdot t d t

=2\left[\tan ^{-1} t \cdot \frac{t^{2}}{2}-\int \frac{1}{1+t^{2}} \cdot \frac{t^{2}}{2} d t\right] \ldots\left\{\int u . v d x=u \int v d x-\int \frac{d u}{d x} \int v d x d x\right\}

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

 

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