Ask & Answer Question

Integrals Ncert Solutions Maths class 12th Maths Class 12th Maths Integrals

Kindly Consider the following

118. $\frac{x}{{(x - 1)}^{2} (x - 2)}$

2 Views | Posted 4 months ago

Asked by Shiksha User

1 Answer

Answered by

Vishal Baghel | Contributor-Level 10

4 months ago

$\frac{x}{{(x - 1)}^{2} (x - 2)} = \frac{A}{(x - 1)} + \frac{B}{{(x - 1)}^{2}} + \frac{C}{x + 2}$

= A(x- 1)(x + 2) + B(x + 2) + C(x- 1)²

= A(x² + 2x-x- 2) + B(x + 2) + C(x² + 1 - 2x)

= A(x² + x- 2) + B(x + 2) + C(x²- 2x + 1)

Comparing the co-efficient we get, A + C = 0 ¾ (1)

A + B - 2C = 1 ¾ (2)

- 2A + 2B + C = 0 ¾ (3)

EQⁿ (3) - 2 ´ eⁿQ. (2),

- 2A + 2B + C (2A + 2B - 4C) = 0 - 2 ´ 1.

=> - 4A + 5C = - 2 ¾ (4)

Eqⁿ (4) + 4 ´ Eqⁿ (1) we get,

- 4A + 5C + 4A + 4C = - 2 + 4 ´ 0

=> 9C = - 2

$\begin{array}{l} \Rightarrow C = - \frac{2}{9} \\ P u t t i n g C = - \frac{2}{9}, i n - - - - - (1) \\ P u t t i n g v a l u e o f A a n d C i n (3) w e g e t, \\ - 2 \times \frac{2}{9} + 2 B - \frac{2}{9} = 0 \\ \Rightarrow 2 B = \frac{2}{9} + \frac{4}{9} \\ \Rightarrow 2 B = \frac{6}{9} : \frac{2}{3} \\ \Rightarrow B = \frac{1}{3} . \end{array}$

<math><mrow><mfrac><mrow><mi>x</mi></mrow><mrow><msup><mrow><mo stretchy="false">(</mo><mi>x</mi><mo>−</mo><mn>1</mn><mo stretchy="false">)</mo></mrow><mrow><mn>2</mn></mrow></msup><mo stretchy="false">(</mo><mi>x</mi><mo>−</mo><mn>2</mn><mo stretchy="false">)</mo></mrow></mfrac><mo>=</mo><mfrac><mrow><mi>A</mi></mrow><mrow><mo stretchy="false">(</mo><mi>x</mi><mo>−</mo><mn>1</mn><mo stretchy="false">)</mo></mrow></mfrac><mo>+</mo><mfrac><mrow><mi>B</mi></mrow><mrow><msup><mrow><mo stretchy="false">(</mo><mi>x</mi><mo>−</mo><mn>1</mn><mo stretchy="false">)</mo></mrow><mrow><mn>2</mn></mrow></msup></mrow></mfrac><mo>+</mo><mfrac><mrow><mi>C</mi></mrow><mrow><mi>x</mi><mo>+</mo><mn>2</mn></mrow></mfrac></mrow></math>= A(x- 1)(x + 2) + B(x + 2) + C(x- 1)2= A(x2 + 2x-x- 2) + B(x + 2) + C(x2 + 1 - 2x)= A(x2 + x- 2) + B(x + 2) + C(x2- 2x + 1)Comparing the co-efficient we get, A + C = 0 ¾ (1)A + B - 2C = 1 ¾ (2)- 2A + 2B + C = 0 ¾ (3)EQn (3) - 2 ´ enQ. (2),- 2A + 2B + C (2A + 2B - 4C) = 0 - 2 ´ 1.=> - 4A + 5C = - 2 ¾ (4)Eqn (4) + 4 ´ Eqn (1) we get,- 4A + 5C + 4A + 4C = - 2 + 4 ´ 0=> 9C = - 2 <math> <mtable columnalign="left"> <mtr> <mtd> <mrow> <mo>⇒</mo> <mi>C</mi> <mo>=</mo> <mo>−</mo> <mfrac> <mrow> <mn>2</mn> </mrow> <mrow> <mn>9</mn> </mrow> </mfrac> </mrow> </mtd> </mtr> <mtr> <mtd> <mrow><mi>P</mi><mi>u</mi><mi>t</mi><mi>t</mi><mi>i</mi><mi>n</mi><mi>g</mi><mtext> </mtext><mi>C</mi><mo>=</mo><mo>−</mo><mfrac><mrow><mn>2</mn></mrow><mrow><mn>9</mn></mrow></mfrac><mo>,</mo><mo> </mo><mi>i</mi><mi>n</mi><mo>−</mo><mo>−</mo><mo>−</mo><mo>−</mo><mo>−</mo> <mrow><mo>(</mo><mrow><mn>1</mn></mrow><mo>)</mo></mrow></mrow> </mtd> </mtr> <mtr> <mtd> <mrow><mi>P</mi><mi>u</mi><mi>t</mi><mi>t</mi><mi>i</mi><mi>n</mi><mi>g</mi> <mi>v</mi><mi>a</mi><mi>l</mi><mi>u</mi><mi>e</mi> <mi>o</mi><mi>f</mi> <mi>A</mi> <mi>a</mi><mi>n</mi><mi>d</mi> <mi>C</mi> <mi>i</mi><mi>n</mi> <mrow><mo>(</mo><mrow><mn>3</mn></mrow><mo>)</mo></mrow> <mi>w</mi><mi>e</mi> <mi>g</mi><mi>e</mi><mi>t</mi><mo>,</mo></mrow> </mtd> </mtr> <mtr> <mtd> <mrow> <mo>−</mo> <mn>2</mn> <mo>×</mo> <mfrac> <mrow> <mn>2</mn> </mrow> <mrow> <mn>9</mn> </mrow> </mfrac> <mo>+</mo> <mn>2</mn> <mi>B</mi> <mo>−</mo> <mfrac> <mrow> <mn>2</mn> </mrow> <mrow> <mn>9</mn> </mrow> </mfrac> <mo>=</mo> <mn>0</mn> </mrow> </mtd> </mtr> <mtr> <mtd> <mrow> <mo>⇒</mo> <mn>2</mn> <mi>B</mi> <mo>=</mo> <mfrac> <mrow> <mn>2</mn> </mrow> <mrow> <mn>9</mn> </mrow> </mfrac> <mo>+</mo> <mfrac> <mrow> <mn>4</mn> </mrow> <mrow> <mn>9</mn> </mrow> </mfrac> </mrow> </mtd> </mtr> <mtr> <mtd> <mrow> <mo>⇒</mo> <mn>2</mn> <mi>B</mi> <mo>=</mo> <mfrac> <mrow> <mn>6</mn> </mrow> <mrow> <mn>9</mn> </mrow> </mfrac> <mo>:</mo> <mfrac> <mrow> <mn>2</mn> </mrow> <mrow> <mn>3</mn> </mrow> </mfrac> </mrow> </mtd> </mtr> <mtr> <mtd> <mrow> <mo>⇒</mo> <mi>B</mi> <mo>=</mo> <mfrac> <mrow> <mn>1</mn> </mrow> <mrow> <mn>3</mn> </mrow> </mfrac> <mo>.</mo> </mrow> </mtd> </mtr> </mtable> </math> <div><div><picture><img src="https://images.shiksha.com/mediadata/images/articles/1742894750phpUWUP5l.jpeg" alt="" width="328" height="279"></picture></div></div>

0 Upvotes 0 Downvotes

Similar Questions for you

The value of $\int_{0}^{π / 4} \frac{x d x}{s i n^{4} (2 x) + c o s^{4} (2 x)}$ equal to

alok kumar singh

$I = \int_{0}^{π / 4} \frac{x d x}{s i n^{4} (2 x) + c o s^{4} (2 x)}$

Let 2x = t then $d x = \frac{1}{2} d t$

$I = \int_{}^{} \frac{\frac{t}{2} \cdot \frac{1}{2} d t}{s i n^{4} t + c o s^{4} t}$

$= \frac{1}{4} \int_{0}^{π / 2} \frac{t d t}{s i n^{4} t + c o s^{4} t} d t$

$I = \frac{1}{4} \int_{0}^{π / 2} \frac{(\frac{π}{2} - t) d t}{s i n^{4} t + c o s^{4} t} d t$

$2 I = \frac{1}{4} \int_{0}^{π / 2} \frac{\frac{π}{2} d t}{s i n^{4} t + c o s^{4} t}$

$2 I = \frac{π}{8} \int_{0}^{π / 2} \frac{s i n^{4} t d t}{t a n^{4} t + 1}$

Let tan t = y then

$2 I = \frac{π}{8} \int_{0}^{\infty} \frac{(1 + y^{2}) d y}{1 + y^{4}}$

$= \frac{π}{8} \int_{0}^{\infty} \frac{1 + \frac{1}{y^{2}}}{y^{2} + \frac{1}{y^{2}} - 2 + 2} d y$

$= \frac{π}{8} \int_{0}^{\infty} \frac{(1 + \frac{1}{y^{2}}) d y}{2 + {(y - \frac{1}{y})}^{2}}$

Let $y - \frac{1}{y} = u$

$2 I = \frac{π}{8} \int_{- \infty}^{\infty} \frac{d u}{2 + u^{2}}$

$= \frac{π}{8 \sqrt[]{2}} {[t a n^{- 1} \frac{4}{\sqrt[]{2}}]}_{- \infty}^{\infty}$

$I = \frac{π^{2}}{16 \sqrt[]{2}}$

If value of $\frac{{(\vec{a} - \vec{b}) \times (\vec{a} - \vec{b} - \vec{c})} . (\vec{a} + 2 \vec{b} - \vec{c})}{[\vec{a} \vec{b} \vec{c}]}$ is k then $\frac{k}{4}$ is

alok kumar singh

$k = \frac{[\vec{a} \vec{b} \vec{c}] + 2 [\vec{a} \vec{b} \vec{c}] + [\vec{a} \vec{b} \vec{c}] - [\vec{a} \vec{b} \vec{c}]}{[\vec{a} \vec{b} \vec{c}]}$

k = 3

If the value of $\underset{x \to 0}{L i m} {s i n^{2} (\frac{π}{2 - 3 x})}^{s e c^{2} (\frac{π}{2 - 5 x})}$ is e^-A then ‘A’ is

alok kumar singh

$\underset{x \to 0}{l i m} {[s i n^{2} (\frac{π}{2 - 3 x})]}^{s e c^{2} (\frac{π}{2 - 5 x})}$

$e^{\underset{x \to 0}{l i m} [s i n^{2} (\frac{π}{2 - 3 x}) - 1] s e c^{2} (\frac{π}{2 - 5 x})}$

$= e^{\underset{x \to 0}{l i m} \frac{- s i n^{2} (\frac{- 3 π x}{2 (2 - 3 x)})}{s i n^{2} (\frac{- 5 π x}{2 (2 - 5 x)})}} = e^{- \frac{9}{25}}$

For real numbers α, β, γ and δ, if ∫ [ (x²-1) + tan⁻¹((x²+1)/x) ] / [ (x⁴+3x²+1)tan⁻¹((x²-1)/x) ] dx = αlogₑ|tan⁻¹((x²+1)/x)| + βtan⁻¹((γ(x²-1))/x) + δtan⁻¹((x²+1)/x) + C where C is an arbitrary constant, then the value of 10(α + βγ + δ) is equal to........

alok kumar singh

The integral is I = ∫ [ (x²-1) + tan? ¹ (x + 1/x)] / [ (x? +3x²+1)tan? ¹ (x+1/x)] dx
This is a complex integral. The provided solution splits it into two parts:
I? = ∫ (x²-1) / [ (x? +3x²+1)tan? ¹ (x+1/x)] dx
I? = ∫ 1 / (x? +3x²+1) dx
The solution proceeds with substitutions which are hard to follow due to OCR quality, but it seems to compare the final result with a given form to find coefficients α, β, γ, δ. The final expression shown is:
10 (α + βγ + δ) = 10 (1 + (1/2√5)*√5 + 1/2) seems incorrect.
The calculation is shown as 10 (1

...more

Kindly considwer the following

alok kumar singh

The problem is to evaluate the integral:
I = ∫? ¹? [x] * e^ [x] / e^ (x-1) dx, where [x] denotes the greatest integer function.

The solution breaks the integral into a sum of integrals over unit intervals:
I = ∑? ∫? ¹ n * e? / e^ (x-1) dx
= ∑? n * e? ∫? ¹ e^ (1-x) dx
= ∑? n * e? [-e^ (1-x)] from n to n+1
= ∑? n * e? [-e? - (-e¹? )]
= ∑? n * e? (e¹? - e? )
= ∑? n * e? * e? (e - 1)
= (e - 1) ∑? n
= (e - 1) * (0 + 1 + 2 + . + 9)
= (e - 1) * (9 * 10 / 2)
= 45 (e - 1)

Get authentic answers from experts, students and alumni that you won't find anywhere else

On Shiksha, get access to

65k Colleges
1.2k Exams
682k Reviews
1800k Answers

Community GuidelinesUser Points System

QuestionsDiscussionsTags

Kindly Consider the following

118. $\frac{x}{{(x - 1)}^{2} (x - 2)}$

1 Answer

Similar Questions for you

Learn more about...

Most viewed information

Most viewed information

Share Your College Life Experience

Didn't find the answer you were looking for?

Need guidance on career and education? Ask our experts

Your Question

Kindly Consider the following 118. x(x−1)2(x−2)

1 Answer

Similar Questions for you

Learn more about...

Most viewed information

Most viewed information

Share Your College Life Experience

Didn't find the answer you were looking for?

Need guidance on career and education? Ask our experts

Your Question

Kindly Consider the following

118. $\frac{x}{{(x - 1)}^{2} (x - 2)}$