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  • Need solution for RD Sharma Maths Class 12 Chapter 18 Indefinite Integrals Excercise 18.32 Question 11

Need solution for RD Sharma Maths Class 12 Chapter 18 Indefinite Integrals Excercise 18.32 Question 11

Answers (1)

Answer : -        \frac{1}{\sqrt{3}} \tan ^{-1}\left(\sqrt{\frac{x^{2}+1}{3}}\right)+C

Hint :-                Use substitution method and special integration formula..

Given :-                \int \frac{x}{\left(x^{2}+4\right) \sqrt{x^{2}+1}} d x

Sol : -             Let \mathrm{I}=\int \frac{x}{\left(x^{2}+4\right) \sqrt{x^{2}+1}} d x                      

                           \begin{array}{r} \text { Put } x^{2}+1=t^{2} \Rightarrow 2 \mathrm{xd} \mathrm{x}=2 \mathrm{t} \mathrm{dt} \rightarrow \mathrm{x} \mathrm{dx}=\mathrm{t} \mathrm{dt} \text { than, } \\\\ \mathrm{I}=\int \frac{1}{\left(t^{2}+3\right) \sqrt{t^{2}}} \mathrm{t} \mathrm{dt}\left(\because x^{2}+1=\mathrm{t}^{2} \rightarrow x^{2}=t^{2}-1\right. \\\\ \left.\rightarrow x^{2}+4=t^{2}-1+4 \rightarrow x^{2}+4=t^{2}+3\right) \end{array}

                       \begin{aligned} &\mathrm{I}=\int \frac{1}{\left(t^{2}+3\right) \cdot t} \mathrm{t} \mathrm{dt} \\ &\mathrm{I}=\int \frac{1}{t^{2}+3} \mathrm{dt} \\ &\mathrm{I}=\int \frac{1}{t^{2}+(\sqrt{3})^{2}} \mathrm{dt} \end{aligned}

                       \begin{aligned} &I=\frac{1}{\sqrt{3}} \tan ^{-1}\left(\frac{t}{\sqrt{2}}\right)+c \quad\left[\because \int \frac{1}{a^{2}+x^{2}} d x=\frac{1}{a} \tan ^{-1}\left(\frac{x}{a}\right)+c\right] \\ &I=\frac{1}{\sqrt{3}} \tan ^{-1}\left(\frac{\sqrt{x^{2}+1}}{\sqrt{3}}\right)+c \\ &I=\frac{1}{\sqrt{3}} \tan ^{-1}\left(\sqrt{\frac{x^{2}+1}{3}}\right)+c \quad\left[\because t=\sqrt{\left.x^{2}+1\right]}\right. \end{aligned}

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