Integration using Partial Fractions MCQ Quiz in मल्याळम - Objective Question with Answer for Integration using Partial Fractions - സൗജന്യ PDF ഡൗൺലോഡ് ചെയ്യുക

Last updated on Apr 23, 2025

നേടുക Integration using Partial Fractions ഉത്തരങ്ങളും വിശദമായ പരിഹാരങ്ങളുമുള്ള മൾട്ടിപ്പിൾ ചോയ്സ് ചോദ്യങ്ങൾ (MCQ ക്വിസ്). ഇവ സൗജന്യമായി ഡൗൺലോഡ് ചെയ്യുക Integration using Partial Fractions MCQ ക്വിസ് പിഡിഎഫ്, ബാങ്കിംഗ്, എസ്എസ്‌സി, റെയിൽവേ, യുപിഎസ്‌സി, സ്റ്റേറ്റ് പിഎസ്‌സി തുടങ്ങിയ നിങ്ങളുടെ വരാനിരിക്കുന്ന പരീക്ഷകൾക്കായി തയ്യാറെടുക്കുക

Latest Integration using Partial Fractions MCQ Objective Questions

Top Integration using Partial Fractions MCQ Objective Questions

Integration using Partial Fractions Question 1:

Evaluate: x+1x23x+2dx

  1. 2log|x1|+3log|x2|+c
  2. 2log|x1|+3log|x2|+C
  3. 2log|x1|+3log|x+2|+c
  4. 2log|x+1|+3log|x2|+c

Answer (Detailed Solution Below)

Option 1 : 2log|x1|+3log|x2|+c

Integration using Partial Fractions Question 1 Detailed Solution

Concept:

  • 1x+adx=log|x+a|+C

Calculation:

To solve: x+1x23x+2dx

The integrand is a proper rational fraction. So, by using the form of partial fraction, we write

⇒ x+1x23x+2=Ax1+Bx2

⇒ x + 1 = (A + B)x - 2A - B

⇒ A + B = 1 and 2A + B = - 1

By solving these equation, we get  A = -2  and  B = 3

⇒ x+1x23x+2=2x1+3x2

⇒ x+1x23x+2dx=2x1dx+3x2dx

⇒ x+1x23x+2dx=2log|x1|+3log|x2|+c

Hence, option 1 is correct.

Integration using Partial Fractions Question 2:

Evaluate: x2+1(x+1)2dx

  1. x2log|x+1|2x+1+C
  2. x+2log|x+1|2x+1+C
  3. x+2log|x+1|+2x+1+C
  4. None of these

Answer (Detailed Solution Below)

Option 1 : x2log|x+1|2x+1+C

Integration using Partial Fractions Question 2 Detailed Solution

Concept:

Partial Fraction:

Factors in the denominator

Corresponding Partial Fraction

(x - a)

Axa

(x – b)2

Axb+B(xb)2

(x - a) (x – b)

A(xa)+B(xb)

(x – c)3

Axc+B(xc)2+C(xc)3

(x – a) (x– a)

A(xa)+Bx+C(x2a)

(ax2 + bx + c)

Ax+B(ax2+bx+c)

 

Calculation:

Here we have to find the value of x2+1(x+1)2dx

As we can see that the given integrand is not a proper rational function, So by division algorithm

x2+1(x+1)2=12x(x+1)2

x2+1(x+1)2dx=dx2x(x+1)2dx ---------(1)

Let's find the value of 2x(x+1)2dx

Let 2x(x+1)2=A(x+1)+B(x+1)2

⇒ 2x = A (x + 1) + B

On equating the coefficients of x in the above equation we get A = 2

On equating the constant terms, we get A + B = 0

∵ A = 2 ⇒ B = - 2

2x(x+1)2=2(x+1)2(x+1)2

2x(x+1)2dx=2dxx+12dx(x+1)2

As we know that,

 dxx=log|x|+C  and xndx=xn+1n+1+C where C is a constant

2x(x+1)2dx=2log|x+1|+2x+1

By substituting the value of 2x(x+1)2dx in equation (1) we get

x2+1(x+1)2dx=x2log|x+1|2x+1+C

Integration using Partial Fractions Question 3:

Evaluate: cosx(1sinx)(2sinx)dx

  1. log|2+sinx1sinx|+C
  2. log|2sinx1sinx|+C
  3. log|2+sinx1+sinx|+C
  4. log|2sinx1+sinx|+C

Answer (Detailed Solution Below)

Option 2 : log|2sinx1sinx|+C

Integration using Partial Fractions Question 3 Detailed Solution

Concept:

Partial Fraction:

Factors in the denominator

Corresponding Partial Fraction

(x - a)

Axa

(x – b)2

Axb+B(xb)2

(x - a) (x – b)

A(xa)+B(xb)

(x – c)3

Axc+B(xc)2+C(xc)3

(x – a) (x2 – a)

A(xa)+Bx+C(x2a)

(ax2 + bx + c)

Ax+B(ax2+bx+c)

 

Calculation:

Here we have to find the value of cosx(1sinx)(2sinx)dx

Let sin x = t and cos x dx = dt

cosx(1sinx)(2sinx)dx=dt(1t)(2t)

Let 1(1t)(2t)=A1t+B2t

⇒ 1 = A (2 - t) + B (1 - t) ----------(1)

By putting t = 1 on both the sides of (1) we get A = 1

By putting t = 2 on both the sides of (1) we get B = - 1

1(1t)(2t)=11t12t

dt(1t)(2t)=dt1tdt2t

As  we know that dxx=log|x|+C  where C is a constant

dt(1t)(2t)=log|1t|+log|2t|+C

=log|2t1t|+C

By substituting sin x = t in the above equation we get,

cosx(1sinx)(2sinx)dx=log|2sinx1sinx|+C

Integration using Partial Fractions Question 4:

Evaluate: dxx2+5x+6=?

  1. logx+2x+3+C
  2. logx+1x+3+C
  3. logx2x+3+C
  4. logx+2x3+C

Answer (Detailed Solution Below)

Option 1 : logx+2x+3+C

Integration using Partial Fractions Question 4 Detailed Solution

Concept:

dxxadxxb=log(xa)log(xb)=logxaxb

Calculation:

Here, we have to find the value of the integrand dxx2+5x+6

dxx2+5x+6=dx(x+2)(x+3)=dxx+2dxx+3

Using the formula, dxxadxxb=log(xa)log(xb)=logxaxb

dxx+2dxx+3=log(x+2)log(x3)=logx+2x+3

dxx2+5x+6=logx+2x+3+C

Hence, the correct option is 1.

Integration using Partial Fractions Question 5:

Evaluate: dxx28x+15=?

  1. 12logx5x+3+C
  2. 12logx5x3+C
  3. 12logx+5x3+C
  4. 12logx2x3+C

Answer (Detailed Solution Below)

Option 2 : 12logx5x3+C

Integration using Partial Fractions Question 5 Detailed Solution

Concept:

dxxadxxb=log(xa)log(xb)=logxaxb

Calculation:

Here, we have to find the value of the integrand dxx28x+15

dxx28x+15=dx(x5)(x3)=12(dxx5dxx3)

Using the formula, dxxadxxb=log(xa)log(xb)=logxaxb

12(dxx5dxx3)=12[log(x5)log(x3)]=12logx5x3

dxx28x+15=12logx5x3+C

Hence, the correct option is 2.

Integration using Partial Fractions Question 6:

Find the value of a + b + d if dxx212x+32=1dlogxaxb+C

  1. 12
  2. 20
  3. 10
  4. 16

Answer (Detailed Solution Below)

Option 4 : 16

Integration using Partial Fractions Question 6 Detailed Solution

Concept:

  • dxax2+bx+c=dx(xb)(xa)=AdxxaBdxxb
  • dxxadxxb=log(xa)log(xb)+C=logxaxb+C

Calculation:

Given: dxx212x+32=1dlogxaxb+C

Using the formula, dxax2+bx+c=dx(xb)(xa)=AdxxaBdxxb

dxx212x+32=dx(x8)(x4)=14(dxx8dxx4)

Using the formula, dxxadxxb=log(xa)log(xb)=logxaxb

14(dxx8dxx4)=14[log(x8)log(x4)]=14logx8x4

dxx212x+32=14logx8x4+C

On comparing the above equation with the given condition dxx212x+32=1dlogxaxb+C we get

⇒ a = 8, b = 4, d = 4..

⇒ a + b + d = 8 + 4 + 4 = 16.

Hence, the correct option is 4. 

Integration using Partial Fractions Question 7:

Evaluate: dxx216=?

  1. 18logx4x+4+C
  2. 18logx4x2+C
  3. 14logx4x+4+C
  4. 12logx4x+4+C

Answer (Detailed Solution Below)

Option 1 : 18logx4x+4+C

Integration using Partial Fractions Question 7 Detailed Solution

Concept:

  • dxax2+bx+c=dx(xb)(xa)=AdxxaBdxxb
  • dxxadxxb=log(xa)log(xb)=logxaxb

Calculation:

Here, we have to evaluate the integral dxx216

Using the formula, dxax2+bx+c=dx(xb)(xa)=AdxxaBdxxb

dxx216=dx(x+4)(x4)=18(dxx4dxx+4)

Using the formula, dxxadxxb=log(xa)log(xb)=logxaxb

18(dxx4dxx+4)=18[log(x4)log(x+4)]=18logx4x+4

dxx216=18logx4x+4+C

Hence, the correct option is 1. 

Integration using Partial Fractions Question 8:

dxx5+x3 = ?

  1. 12x2logx+12log(x2+1) + C
  2. 12x2logx+12log(x2+1) + C
  3. 12x2+logx12log(x2+1) + C
  4. None of these

Answer (Detailed Solution Below)

Option 1 : 12x2logx+12log(x2+1) + C

Integration using Partial Fractions Question 8 Detailed Solution

Concept:

Integral property:

  • ∫ xn dx = xn+1n+1+ C ; n ≠ -1
  • 1xdx=lnx + C

 

 

 

Calculation:

I = dxx5+x3

I = dxx3(x2+1) 

By partial fraction

I = xx2+11x+1x3 dx

I = xx2+1dxdxx+dxx3

Let x2 + 1 = t ⇒ 2x dx = dt

I = dt2tlnx+x3dx+C

I = 12lntlnx+x22+C

Putting the value of t

I = 12ln(x2+1)lnx12x2+C

Integration using Partial Fractions Question 9:

Find the value of the  xdxx2+3x+2

  1. log(x2)2x+1+C
  2. log(x+2)2x+1+C
  3. log(x+3)2x+1+C
  4. log(x+2)2x1+C

Answer (Detailed Solution Below)

Option 2 : log(x+2)2x+1+C

Integration using Partial Fractions Question 9 Detailed Solution

Concept:

  • dxxadxxb=log(xa)log(xb)=logxaxb

Calculation:

Here, we have to find the value of the integrand xdxx2+3x+2

xdxx2+3x+2=dx(x+1)(x+2)=dxx+1+2dxx+2

Using the formula, dxxadxxb=log(xa)log(xb)=logxaxb

dxx+1+2dxx+2=[log(x+1)+2log(x+2)]=log(x+2)2x+1

xdxx2+3x+2=log(x+2)2x+1+C

Hence, the correct option is 2. 

Integration using Partial Fractions Question 10:

If dx(x+2)(x2+1) = a log |1 + x2| + b tan−1x + 15 log |x + 2| + C, then

  1. a = 110, b = 25
  2. a = 110, b = 25
  3. a = 110, b = 25
  4. a = 110, b = 25

Answer (Detailed Solution Below)

Option 3 : a = 110, b = 25

Integration using Partial Fractions Question 10 Detailed Solution

Concept: 

f(x)f(x) dx=log|f(x)|+C

dx1+x2=tan1x+C

Calculation: 

Given:  dx(x+2)(x2+1) = a log |1 + x2| + b tan−1x + 15 log |x + 2| + C

Let 1(x+2)(x2+1)=Ax+2+Bx+Cx2+1

Then 1(x+2)(x2+1)=A(x2+1)+(Bx+C)(x+2)(x+2)(x2+1)

1=A(x2+1)+(Bx+C)(x+2)

On comparing the coefficients of x2 on both sides we get

0 = A + B --- (i)

On comparing the coefficients of x on both sides we get

0 = 2B + C --- (ii)

On comparing the Constants on both sides we get

1 = A + 2C --- (iii)

2 × (i) - (ii) ⇒ 2A - C = 0 --- (iv)

(iii) + 2 × (iv) ⇒ 5A = 1 ⇒ A = 15

From (iv), C = 25

From (ii), B = 15

Therefore 1(x+2)(x2+1)=15x+2+15x+25x2+1

Then dx(x+2)(x2+1)=[15x+2+15x+25x2+1] dx

dx(x+2)(x2+1)=151x+2 dx15xx2+1 dx+251x2+1 dx

dx(x+2)(x2+1)=15log|x+2|110log|x2+1|+25tan1x+C

On comparing with  dx(x+2)(x2+1) = a log |1 + x2| + b tan−1x + 15 log |x + 2| + C we get

a = 110 and b = 25

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