# Class 12 Maths MCQ – Integration by Partial Fractions

This set of Class 12 Maths Chapter 7 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) $$\frac{A}{(x-a)} + \frac{B}{(x-b)}$$

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.

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

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) $$Log \left|\frac{x+1}{x-2}\right|+ C$$

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
= $$Log \left|\frac{x+1}{x+2}\right|+ C$$.

4. An improper integration fraction is reduced to proper fraction by _____
a) multiplication
b) division
d) subtraction

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

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

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) $$\frac{A}{(x-a)} – \frac{B}{(x-a)}$$

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.

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) $$\frac{3}{5}log|x+2| + \frac{1}{5}log|x^2+1|+\frac{1}{5} tan^{-1}x+C$$

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

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.

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

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.

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