Mathematics Questions and Answers – Integration by Partial Fractions

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This set of Mathematics Multiple Choice Questions & Answers (MCQs) focuses on “Integration by Partial Fractions”.

1. What form of rational function \(\frac{px+q}{(x-a)(x-b)}\), a≠b represents?
a) \(\frac{A}{(x-a)}\)
b) \(\frac{B}{(x-b)}\)
c) \(\frac{A+B}{(x-a)(x-b)}\)
d) \(\frac{A}{(x-a)} + \frac{B}{(x-b)}\)
View Answer

Answer: d
Explanation: The given function \(\frac{px+q}{(x-a)(x-b)}\), a≠b can also be written as
\(\frac{A}{(x-a)} + \frac{B}{(x-b)}\) and is further used to solve integration by partial fractions numerical.
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2. Find \(\int \frac{x^2+1}{x^2-5x+6} dx\).
a) x – 5log|x-2| + 10log|x-3|+C
b) x – 3log|x-2| + 5log|x-3|+C
c) x – 10log|x-2| + 5log|x-3|+C
d) x – 5log|x-5| + 10log|x-10|+C
View Answer

Answer: a
Explanation: As it is not proper rational function, we divide numerator by denominator and get
\(\frac{x^2+1}{x^2-5x+6} = 1-\frac{5x-5}{x^2-5x+6} = 1+\frac{5x-5}{(x-2)(x-3)}\)
Let \(\frac{5x-5}{(x-2)(x-3)}=\frac{A}{(x-2)} + \frac{B}{(x-3)}\)
So that, 5x–5 = A(x-3) + B(x-2)
Now, equating coefficients of x and constant on both sides, we get A + B = 5 and 3A + 2B = 5. Solving these equations, we get A=-5 and B=10.
Therefore, \(\frac{x^2+1}{x^2-5x+6} = 1 – \frac{5}{(x-2)} + \frac{10}{(x-3)}\).
\(\int \frac{x^2+1}{x^2-5x+6} dx = \int dx – 5\int \frac{dx}{(x-2)} + 10\int \frac{dx}{(x-3)}\).
= x – 5log|x-2| + 10log|x-3|+C

3. Find \(\int \frac{dx}{(x+1)(x+2)}\).
a) \(Log \left|\frac{x+1}{x+2}\right|+ C\)
b) \(Log \left|\frac{x-1}{x+2}\right|+ C\)
c) \(Log \left|\frac{x+2}{x+1}\right|+ C\)
d) \(Log \left|\frac{x+1}{x-2}\right|+ C\)
View Answer

Answer: a
Explanation: It is a proper rational function. Therefore,
\(\frac{1}{(x+1)(x+2)} = \frac{A}{(x+1)} + \frac{B}{(x+2)}\)
Where real numbers are determined, 1 = A(x+2) + B(x+1), Equating coefficients of x and the constant term, we get A+B = 0 and 2A+B = 1. Solving it we get A=1, and B=-1.
Thus, it simplifies to, \(\frac{1}{(x+1)} + \frac{-1}{(x+2)} = \int \frac{dx}{(x+1)} – \int \frac{dx}{(x+2)}\).
= log|x+1| – log|x+2| + C
= \(Log \left|\frac{x+1}{x+2}\right|+ C\).
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4. An improper integration fraction is reduced to proper fraction by _____
a) multiplication
b) division
c) addition
d) subtraction
View Answer

Answer: b
Explanation: An improper integration factor can be reduced to proper fraction by division, i.e., if the numerator and denominator have same degree, then they must be divided in order to reduce it to proper fraction.

5. \(\int \frac{dx}{x(x^2+1)}\) equals ______
a) \(log|x| – \frac{1}{2} log(x^2+1)\) + C
b) \(log|x| + \frac{1}{2} log(x^2+1)\) + C
c) –\(log|x| + \frac{1}{2} log(x^2+1)\) + C
d) \(\frac{1}{2} log|x| + log(x^2+1)\) + C
View Answer

Answer: a
Explanation: We know that \(\int \frac{dx}{x(x^2+1)} = \frac{A}{x} + \frac{Bx+C}{x^2+1}\)
By simplifying it we get, \(\int \frac{dx}{x(x^2+1)}=\frac{(A+B) x^2+Cx+A}{x(x^2+1)}\)
Now equating the coefficients we get A = 0, B = 0, C=1.
\(\int \frac{dx}{x(x^2+1)} = \int \frac{dx}{x} + \int \frac{-xdx}{(x^2+1)}\)
Therefore after integrating we get \(log|x| – \frac{1}{2} log(x^2+1)\) + C.
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6. \(\int \frac{dx}{(x^2-9)}\) equals ______
a) \(\frac{1}{6} log \frac{x+3}{x-3}\) + C
b) \(\frac{1}{6} log \frac{x-3}{x+3}\) + C
c) \(\frac{1}{5} log \frac{x+3}{x-3}\) + C
d) \(\frac{1}{3} log \frac{x+3}{x-3}\) + C
View Answer

Answer: b
Explanation: \(\int \frac{dx}{(x^2-9)}=\frac{A}{(x-3)} + \frac{B}{(x+3)}\)
By simplifying, it we get \(\frac{A(x+3)+B(x-3)}{(x^2-9)} = \frac{(A+B)x+3A-3B}{(x^2-9)}\)
By solving the equations, we get, A+B=0 and 3A-3B=1
By solving these 2 equations, we get values of A=1/6 and B=-1/6.
Now by putting values in the equation and integrating it we get value,
\(\frac{1}{6} log (\frac{x-3}{x+3})\) + C.

7. Which form of rational function \(\frac{px+q}{(x-a)^2}\) represents?
a) \(\frac{A}{(x-a)} + \frac{B}{(x-a)^2}\)
b) \(\frac{A}{(x-a)^2} + \frac{B}{(x-a)}\)
c) \(\frac{A}{(x-a)} – \frac{B}{(x-a)^2}\)
d) \(\frac{A}{(x-a)} – \frac{B}{(x-a)}\)
View Answer

Answer: a
Explanation: It is a form of the given partial fraction \(\frac{px+q}{(x-a)^2}\) which can also be written as
\(\frac{A}{(x-a)} + \frac{B}{(x-a)^2}\) and is further used to solve integration by partial fractions numerical.
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8. \(\int \frac{(x^2+x+1)dx}{(x+2)(x^2+1)}\) equals ______
a) \(\frac{3}{5}log|x+2| + \frac{1}{5}log|x^2+1|+\frac{1}{5} tan^{-1}x+5C\)
b) \(\frac{3}{5}log|x+2| + \frac{1}{5}log|x^2+1|+\frac{1}{6} tan^{-1}x+C\)
c) \(\frac{3}{5}log|x+2| + \frac{1}{6}log|x^2+1|+\frac{1}{6} tan^{-1}x+C\)
d) \(\frac{3}{5}log|x+2| + \frac{1}{5}log|x^2+1|+\frac{1}{5} tan^{-1}x+C\)
View Answer

Answer: d
Explanation: \(\int \frac{(x^2+x+1)dx}{(x+2)(x^2+1)} = \frac{A}{(x+2)} + \frac{Bx+C}{(x^2+1)}\)
Now equating, (x2+x+1) = A (x2+1) + (Bx+C) (x+2)
After equating and solving for coefficient we get values,
A=\(\frac{3}{5}\), B=\(\frac{2}{5}\), C=\(\frac{1}{5}\), now putting these values in the equation we get,
\(\int \frac{(x^2+x+1)dx}{(x+2)(x^2+1)} = \frac{3}{5} \int \frac{dx}{(x+2)} + \frac{1}{5} \int \frac{2xdx}{(x^2+1)} + \frac{1}{5} \int \frac{dx}{(x^2+1)}\)
Hence it comes, \(\frac{3}{5} log|x+2| + \frac{1}{5} log|x^2+1|+\frac{1}{5}tan^{-1}x+C\)

9. Identify the type of the equation (x+1)2.
a) Linear equation
b) Cubic equation
c) Identity
d) Imaginary
View Answer

Answer: c
Explanation: As it represents the identity (b+a)2 it satisfies the identity (b+a)2 = (a2 + b2 +2ab) and is not linear, cubic or an imaginary equation so the correct option is Identity Equation.
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10. For the given equation (x+2) (x+4) = x2 + 6x + 8, how many values of x satisfies this equation?
a) Two values of x
b) One value of x
c) All value of x
d) No value of x
View Answer

Answer: c
Explanation: If we solve the L.H.S. (Left Hand Side) of the equation, we get the following value.
(x+2) (x+4) = x2 + 4x + 2x + 8 = x2 + 6x + 8.
This value is same as the R.H.S. (Right Hand Side).
So, all the values of x satisfy the equality.

Sanfoundry Global Education & Learning Series – Mathematics – Class 12.

To practice all areas of Mathematics, here is complete set of 1000+ Multiple Choice Questions and Answers.

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Manish Bhojasia, a technology veteran with 20+ years @ Cisco & Wipro, is Founder and CTO at Sanfoundry. He lives in Bangalore, and focuses on development of Linux Kernel, SAN Technologies, Advanced C, Data Structures & Alogrithms. Stay connected with him at LinkedIn.

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