Browse by Stream
QnA
Home
QnA
Go To Your Study Dashboard

Get Answers to all your Questions

header-bg qa
  • Home
  • School
  • Need solution for RD Sharma Maths Class 12 Chapter 18 Indefinite Integrals Excercise 18.32 Question 8

Need solution for RD Sharma Maths Class 12 Chapter 18 Indefinite Integrals Excercise 18.32 Question 8

Answers (1)

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

Hint :-                Use substitution and special integration formula.

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

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

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

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

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

\begin{aligned} &=-\int \frac{\frac{1}{t^{2}}}{\frac{1}{t}\sqrt{\frac{1+t^{2}+2t+t^{2}}{t^{2}}}}dt \\ &=-\int \frac{\frac{1}{t^{2}}}{\frac{11}{t} \cdot \frac{1}{t} \sqrt{2 t^{2}+2 t+1}} d t \\ &=-\int \frac{\frac{1}{t^{2}}}{\frac{1}{t^{2}} \sqrt{2\left(t^{2}+t+\frac{1}{2}\right)}} d t \end{aligned}

        \begin{aligned} &=-\int \frac{1}{\sqrt{2}} \cdot \frac{1}{\sqrt{t^{2}+2 \cdot t \cdot \frac{1}{2}+\left(\frac{1}{2}\right)^{2}-\left(\frac{1}{2}\right)^{2}+\frac{1}{2}}} d t \\ &=-\frac{1}{\sqrt{2}} \int \frac{1}{\sqrt{\left(t+\frac{1}{2}\right)^{2}-\frac{1}{4}+\frac{1}{2}}} d t&\quad\left[\because a^{2}+b^{2}+2 a b=(a+b)^{2}\right] \\ \end{aligned}

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

 

\Rightarrow dt=du than

 

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

Posted by

infoexpert27

View full answer

Crack CUET with india's "Best Teachers"

  • HD Video Lectures
  • Unlimited Mock Tests
  • Faculty Support
cuet_ads

Similar Questions

Trending Articles/News

anam-khan

PM Modi's Pariksha Pe Charcha 2025 to be held next month, registrations open till January 14
anam-khan

CBSE Board Exams 2025: Online portal for availing exemptions by CWSN students opens
anam-khan

Board Exams: States agree to equivalence; no question paper ‘jumbling’ from next year, says PARAKH CEO

Latest Question