Browse by Stream
QnA
Home
QnA
Go To Your Study Dashboard

Get Answers to all your Questions

header-bg qa
  • Home
  • School
  • Please Solve RD Sharma Class 12 Chapter 18 Indefinite Integrals Exercise18.31 Question 8 Maths Textbook Solution.

Please Solve RD Sharma Class 12 Chapter 18 Indefinite Integrals Exercise18.31 Question 8 Maths Textbook Solution.

Answers (1)

Answer: The required value of the integral is,

\frac{1}{3} \tan ^{-1}\left(\frac{x^{2}-1}{3 x}\right)+c

Given:

\int \frac{x^{2}+1}{x^{4}+7 x^{2}+1} d x

Hint: The given integral function can be converted into

\int \frac{d x}{x^{2}+a^{2}}=\frac{1}{a} \tan ^{-1} \frac{x}{a}+c

Solution: Rewrite the given equation as.

\int \frac{x^{2}+1}{x^{4}+7 x^{2}+1} d x

Divide numerator and denominator by x^{2}

\begin{aligned} I &=\int \frac{1+\frac{1}{x^{2}}}{x^{2}+\frac{1}{x^{2}}+7} d x \\ &=\int \frac{1+\frac{1}{x^{2}}}{x^{2}+\frac{1}{x^{2}}+7+2-2} dx \\ &=\int \frac{1+\frac{1}{x^{2}}}{\left(x^{2}+\frac{1}{x^{2}}-2\right)+9} d x \\ &=\int \frac{1+\frac{1}{x^{2}}}{\left(x-\frac{1}{x}\right)^{2}+9} d x \end{aligned}

let(x-\frac{1}{x})=t\Rightarrow (1+\frac{1}{x^{2}})dx=dt

Evaluating the obtained integration.

\begin{aligned} &I=\int \frac{d t}{t^{2}+3^{2}} \\ &I=\frac{1}{3} \tan -1\left(\frac{t}{3}\right)+c \\ &I=\frac{1}{3} \tan ^{-1}\left(\frac{\left.x-\frac{1}{x}\right)+c}{3}\right)+c \\ &I=\frac{1}{3} \tan ^{-1}\left(\frac{x^{2}-1}{3 x}\right)+c &\text { (Substituting } t \text { as }\left(x-\frac{1}{x}\right) \text { ) } \\ \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

Board Exam Time Table 2025 Live: UP Board exams from February 24; CBSE, BSEB date sheet soon
anam-khan

CBSE Date Sheet 2025 Live: Class 10, 12 board exam dates soon; syllabus, sample papers
anam-khan

Extra time in CBSE exams for students with type 1 diabetes: Kerala SHRC seeks report on plea

Latest Question