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  • Need solution for RD Sharma Maths Class 12 Chapter 18 Indefinite Integrals Excercise 18.32 Question 9

Need solution for RD Sharma Maths Class 12 Chapter 18 Indefinite Integrals Excercise 18.32 Question 9

Answers (1)

Answer : -       -\log \left|\left(\frac{1}{x+1}-\frac{1}{2}\right)+\sqrt{\left(\frac{1}{x+1}-\frac{1}{2}\right)^{2}+\frac{3}{4}}\right|+C

Hint :-                Use substitution method and special integration formula to solve this integration                                                                                

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

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

                        Put,  x + 1 = \frac{1}{t}

\Rightarrow dx=-\frac{1}{t^{2}}dt   then,

    \begin{aligned} &\mathrm{I}=\int \frac{1}{\frac{1}{t} \sqrt{\left(\frac{1}{t}-1\right)^{2}+\left(\frac{1}{t}-1\right)+1}}\left(-\frac{1}{t^{2}}\right) d t\left(\because \mathrm{x}=\frac{1}{t}-1\right) \\ &=-\int \frac{\frac{1}{t^{2}}}{\frac{1}{t} \sqrt{\frac{1}{t^{2}}+1-\frac{2}{t}+\frac{1}{t}-1+1}} d t \quad \because(a-b)^{2}=a^{2}-2 a b+b^{2} \\ &=-\int \frac{\frac{1}{t^{2}}}{\frac{1}{t} \sqrt{t^{2}-\frac{1}{t}+1}} d t \end{aligned}     

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

 

               \begin{aligned} &=-\int \frac{1}{\sqrt{\left(t-\frac{1}{2}\right)^{2}-\frac{1}{4}+1}} d t \\ &=-\int \frac{1}{\sqrt{\left(t-\frac{1}{2}\right)^{2}-\left(\frac{1-4}{4}\right)}} d t \end{aligned}

 

       \begin{aligned} &=-\int \frac{1}{\sqrt{\left(t-\frac{1}{2}\right)^{2}-\left(\frac{-3}{4}\right)}} d t \\ &=-\int \frac{1}{\sqrt{\left(t-\frac{1}{2}\right)^{2}+\left(\frac{3}{4}\right)}} d t \\ &=-\int \frac{1}{\sqrt{\left(t-\frac{1}{2}\right)^{2}+\left(\frac{\sqrt{3}}{2}\right)^{2}}} d t \end{aligned}

Put,t-\frac{1}{2}=u

\Rightarrow dt=du than,

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

 \begin{aligned} &=-\log \left|\left(t-\frac{1}{2}\right)+\sqrt{\left(t-\frac{1}{2}\right)^{2}+\frac{3}{4}}\right|+\mathrm{c} \quad\left(\because \mathrm{u}=\mathrm{t}-\frac{1}{2}\right) \\ &=-\log \left|\left(\frac{1}{x+1}-\frac{1}{2}\right)+\sqrt{\left(\frac{1}{x+1}-\frac{1}{2}\right)^{2}+\frac{3}{4}}\right| \mid+c \quad\left[\because \mathrm{t}=\frac{1}{x+1}\right] \end{aligned}

 

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