Complete solutions for Class 12 Maths (NCERT) Exercise 7.5. Master the technique of Partial Fraction Decomposition to integrate rational functions with linear, repeated, and irreducible quadratic factors. Includes advanced substitution problems like $\int \frac{1}{x(x^n + 1)} dx$ and $\int \frac{1}{e^x – 1} dx$.
This exercise involves integrating rational functions by decomposing them into simpler fractions using the method of Partial Fraction Decomposition.
Table of Contents
Part 1: Linear Non-Repeated Factors (1-4)
1. $\int \frac{x}{(x + 1)(x + 2)} dx$
Decomposition form:
$$\frac{x}{(x + 1)(x + 2)} = \frac{A}{x + 1} + \frac{B}{x + 2}$$
$$x = A(x + 2) + B(x + 1)$$
- Put $x = -1$: $-1 = A(-1 + 2) + 0 \implies A = -1$.
- Put $x = -2$: $-2 = 0 + B(-2 + 1) \implies B = 2$.$$\int \left( \frac{-1}{x + 1} + \frac{2}{x + 2} \right) dx = -\log|x + 1| + 2 \log|x + 2| + C$$$$= \mathbf{\log \left| \frac{(x + 2)^2}{x + 1} \right| + C}$$
2. $\int \frac{1}{x^2 – 9} dx = \int \frac{1}{(x – 3)(x + 3)} dx$
Decomposition form:
$$\frac{1}{(x – 3)(x + 3)} = \frac{A}{x – 3} + \frac{B}{x + 3}$$
$$1 = A(x + 3) + B(x – 3)$$
- Put $x = 3$: $1 = A(6) \implies A = 1/6$.
- Put $x = -3$: $1 = B(-6) \implies B = -1/6$.$$\int \left( \frac{1/6}{x – 3} – \frac{1/6}{x + 3} \right) dx = \frac{1}{6} (\log|x – 3| – \log|x + 3|) + C$$$$= \mathbf{\frac{1}{6} \log \left| \frac{x – 3}{x + 3} \right| + C}$$(This is a standard formula, $\int \frac{dx}{x^2 – a^2} = \frac{1}{2a} \log\left|\frac{x – a}{x + a}\right| + C$.)
3. $\int \frac{3x – 1}{(x – 1)(x – 2)(x – 3)} dx$
Decomposition form:
$$\frac{3x – 1}{(x – 1)(x – 2)(x – 3)} = \frac{A}{x – 1} + \frac{B}{x – 2} + \frac{C}{x – 3}$$
$$3x – 1 = A(x – 2)(x – 3) + B(x – 1)(x – 3) + C(x – 1)(x – 2)$$
- Put $x = 1$: $3(1) – 1 = A(-1)(-2) \implies 2 = 2A \implies A = 1$.
- Put $x = 2$: $3(2) – 1 = B(1)(-1) \implies 5 = -B \implies B = -5$.
- Put $x = 3$: $3(3) – 1 = C(2)(1) \implies 8 = 2C \implies C = 4$.$$\int \left( \frac{1}{x – 1} – \frac{5}{x – 2} + \frac{4}{x – 3} \right) dx$$$$= \mathbf{\log|x – 1| – 5 \log|x – 2| + 4 \log|x – 3| + C}$$
4. $\int \frac{x}{(x – 1)(x – 2)(x – 3)} dx$
Decomposition form:
$$\frac{x}{(x – 1)(x – 2)(x – 3)} = \frac{A}{x – 1} + \frac{B}{x – 2} + \frac{C}{x – 3}$$
$$x = A(x – 2)(x – 3) + B(x – 1)(x – 3) + C(x – 1)(x – 2)$$
- Put $x = 1$: $1 = A(-1)(-2) \implies 1 = 2A \implies A = 1/2$.
- Put $x = 2$: $2 = B(1)(-1) \implies B = -2$.
- Put $x = 3$: $3 = C(2)(1) \implies 3 = 2C \implies C = 3/2$.$$\int \left( \frac{1/2}{x – 1} – \frac{2}{x – 2} + \frac{3/2}{x – 3} \right) dx$$$$= \mathbf{\frac{1}{2} \log|x – 1| – 2 \log|x – 2| + \frac{3}{2} \log|x – 3| + C}$$
Part 2: Quadratic Non-factorable Factors (5-6)
5. $\int \frac{2x}{x^2 + 3x + 2} dx = \int \frac{2x}{(x + 1)(x + 2)} dx$
Decomposition form:
$$\frac{2x}{(x + 1)(x + 2)} = \frac{A}{x + 1} + \frac{B}{x + 2}$$
$$2x = A(x + 2) + B(x + 1)$$
- Put $x = -1$: $2(-1) = A(-1 + 2) \implies -2 = A$.
- Put $x = -2$: $2(-2) = B(-2 + 1) \implies -4 = -B \implies B = 4$.$$\int \left( \frac{-2}{x + 1} + \frac{4}{x + 2} \right) dx = -2 \log|x + 1| + 4 \log|x + 2| + C$$$$= \mathbf{4 \log|x + 2| – 2 \log|x + 1| + C}$$
6. $\int \frac{1}{x(x^2 – 1)} dx = \int \frac{1}{x(x – 1)(x + 1)} dx$
Decomposition form:
$$\frac{1}{x(x – 1)(x + 1)} = \frac{A}{x} + \frac{B}{x – 1} + \frac{C}{x + 1}$$
$$1 = A(x – 1)(x + 1) + B(x)(x + 1) + C(x)(x – 1)$$
- Put $x = 0$: $1 = A(-1)(1) \implies A = -1$.
- Put $x = 1$: $1 = B(1)(2) \implies B = 1/2$.
- Put $x = -1$: $1 = C(-1)(-2) \implies 1 = 2C \implies C = 1/2$.$$\int \left( \frac{-1}{x} + \frac{1/2}{x – 1} + \frac{1/2}{x + 1} \right) dx$$$$= -\log|x| + \frac{1}{2} \log|x – 1| + \frac{1}{2} \log|x + 1| + C$$$$= \mathbf{-\log|x| + \frac{1}{2} \log|x^2 – 1| + C}$$
Part 3: Repeated and Mixed Factors (7-15)
7. $\int \frac{x}{(x^2 + 1)(x – 1)} dx$
Decomposition form:
$$\frac{x}{(x – 1)(x^2 + 1)} = \frac{A}{x – 1} + \frac{Bx + C}{x^2 + 1}$$
$$x = A(x^2 + 1) + (Bx + C)(x – 1)$$
- Put $x = 1$: $1 = A(1 + 1) \implies A = 1/2$.
- Comparing coefficients of $x^2$: $0 = A + B \implies B = -A = -1/2$.
- Comparing constant terms: $0 = A – C \implies C = A = 1/2$.$$\int \left( \frac{1/2}{x – 1} + \frac{-1/2 x + 1/2}{x^2 + 1} \right) dx = \frac{1}{2} \int \frac{1}{x – 1} dx – \frac{1}{2} \int \frac{x}{x^2 + 1} dx + \frac{1}{2} \int \frac{1}{x^2 + 1} dx$$
- Term 1: $\frac{1}{2} \log|x – 1|$.
- Term 2: Let $u = x^2 + 1$, $du = 2x dx$. $-\frac{1}{4} \int \frac{du}{u} = -\frac{1}{4} \log|x^2 + 1|$.
- Term 3: $\frac{1}{2} \tan^{-1} x$.$$= \mathbf{\frac{1}{2} \log|x – 1| – \frac{1}{4} \log(x^2 + 1) + \frac{1}{2} \tan^{-1} x + C}$$
8. $\int \frac{x}{(x – 1)^2 (x + 2)} dx$
Decomposition form:
$$\frac{x}{(x – 1)^2 (x + 2)} = \frac{A}{x – 1} + \frac{B}{(x – 1)^2} + \frac{C}{x + 2}$$
$$x = A(x – 1)(x + 2) + B(x + 2) + C(x – 1)^2$$
- Put $x = 1$: $1 = 0 + B(3) + 0 \implies B = 1/3$.
- Put $x = -2$: $-2 = 0 + 0 + C(-3)^2 \implies -2 = 9C \implies C = -2/9$.
- Put $x = 0$: $0 = A(-2) + B(2) + C(1) \implies 0 = -2A + 2(1/3) – 2/9 \implies 2A = 2/3 – 2/9 = 4/9 \implies A = 2/9$.$$\int \left( \frac{2/9}{x – 1} + \frac{1/3}{(x – 1)^2} – \frac{2/9}{x + 2} \right) dx$$$$= \frac{2}{9} \log|x – 1| + \frac{1}{3} \left(-\frac{1}{x – 1}\right) – \frac{2}{9} \log|x + 2| + C$$$$= \mathbf{\frac{2}{9} \log \left| \frac{x – 1}{x + 2} \right| – \frac{1}{3(x – 1)} + C}$$
9. $\int \frac{3x + 5}{x^3 – x^2 – x + 1} dx = \int \frac{3x + 5}{(x – 1)^2 (x + 1)} dx$
Denominator: $x^2(x – 1) – 1(x – 1) = (x^2 – 1)(x – 1) = (x – 1)(x + 1)(x – 1) = (x – 1)^2(x + 1)$.
Decomposition form:
$$\frac{3x + 5}{(x – 1)^2 (x + 1)} = \frac{A}{x – 1} + \frac{B}{(x – 1)^2} + \frac{C}{x + 1}$$
$$3x + 5 = A(x – 1)(x + 1) + B(x + 1) + C(x – 1)^2$$
- Put $x = 1$: $3(1) + 5 = 0 + B(2) + 0 \implies 8 = 2B \implies B = 4$.
- Put $x = -1$: $3(-1) + 5 = 0 + 0 + C(-2)^2 \implies 2 = 4C \implies C = 1/2$.
- Put $x = 0$: $5 = A(-1) + B(1) + C(1) \implies 5 = -A + 4 + 1/2 \implies A = -1/2$.$$\int \left( -\frac{1/2}{x – 1} + \frac{4}{(x – 1)^2} + \frac{1/2}{x + 1} \right) dx$$$$= -\frac{1}{2} \log|x – 1| + 4 \left(-\frac{1}{x – 1}\right) + \frac{1}{2} \log|x + 1| + C$$$$= \mathbf{\frac{1}{2} \log \left| \frac{x + 1}{x – 1} \right| – \frac{4}{x – 1} + C}$$
10. $\int \frac{2x – 3}{(x^2 – 1)(2x + 3)} dx = \int \frac{2x – 3}{(x – 1)(x + 1)(2x + 3)} dx$
Decomposition form:
$$\frac{2x – 3}{(x – 1)(x + 1)(2x + 3)} = \frac{A}{x – 1} + \frac{B}{x + 1} + \frac{C}{2x + 3}$$
$$2x – 3 = A(x + 1)(2x + 3) + B(x – 1)(2x + 3) + C(x – 1)(x + 1)$$
- Put $x = 1$: $2(1) – 3 = A(2)(5) \implies -1 = 10A \implies A = -1/10$.
- Put $x = -1$: $2(-1) – 3 = B(-2)(1) \implies -5 = -2B \implies B = 5/2$.
- Put $x = -3/2$: $2(-3/2) – 3 = C(-5/2)(-1/2) \implies -6 = 5/4 C \implies C = -24/5$.$$\int \left( \frac{-1/10}{x – 1} + \frac{5/2}{x + 1} – \frac{24/5}{2x + 3} \right) dx$$$$= -\frac{1}{10} \log|x – 1| + \frac{5}{2} \log|x + 1| – \frac{24}{5} \frac{\log|2x + 3|}{2} + C$$$$= \mathbf{-\frac{1}{10} \log|x – 1| + \frac{5}{2} \log|x + 1| – \frac{12}{5} \log|2x + 3| + C}$$
11. $\int \frac{5x}{(x + 1)(x^2 – 4)} dx = \int \frac{5x}{(x + 1)(x – 2)(x + 2)} dx$
Decomposition form:
$$\frac{5x}{(x + 1)(x – 2)(x + 2)} = \frac{A}{x + 1} + \frac{B}{x – 2} + \frac{C}{x + 2}$$
$$5x = A(x – 2)(x + 2) + B(x + 1)(x + 2) + C(x + 1)(x – 2)$$
- Put $x = -1$: $-5 = A(-3)(1) \implies -5 = -3A \implies A = 5/3$.
- Put $x = 2$: $10 = B(3)(4) \implies 10 = 12B \implies B = 5/6$.
- Put $x = -2$: $-10 = C(-1)(-4) \implies -10 = 4C \implies C = -5/2$.$$\int \left( \frac{5/3}{x + 1} + \frac{5/6}{x – 2} – \frac{5/2}{x + 2} \right) dx$$$$= \mathbf{\frac{5}{3} \log|x + 1| + \frac{5}{6} \log|x – 2| – \frac{5}{2} \log|x + 2| + C}$$
12. $\int \frac{x^3 + x + 1}{x^2 – 1} dx$
Since the degree of the numerator (3) is greater than the degree of the denominator (2), use long division first.
$$\frac{x^3 + x + 1}{x^2 – 1} = x + \frac{2x + 1}{x^2 – 1}$$
We integrate: $\int x dx + \int \frac{2x + 1}{(x – 1)(x + 1)} dx$.
For the partial fraction part:
$$\frac{2x + 1}{(x – 1)(x + 1)} = \frac{A}{x – 1} + \frac{B}{x + 1}$$
$$2x + 1 = A(x + 1) + B(x – 1)$$
- Put $x = 1$: $3 = 2A \implies A = 3/2$.
- Put $x = -1$: $-1 = -2B \implies B = 1/2$.$$\int \left( x + \frac{3/2}{x – 1} + \frac{1/2}{x + 1} \right) dx$$$$= \mathbf{\frac{x^2}{2} + \frac{3}{2} \log|x – 1| + \frac{1}{2} \log|x + 1| + C}$$
13. $\int \frac{2}{(1 – x)(1 + x^2)} dx$
Decomposition form:
$$\frac{2}{(1 – x)(1 + x^2)} = \frac{A}{1 – x} + \frac{Bx + C}{1 + x^2}$$
$$2 = A(1 + x^2) + (Bx + C)(1 – x)$$
- Put $x = 1$: $2 = A(2) \implies A = 1$.
- Comparing coefficients of $x^2$: $0 = A – B \implies B = A = 1$.
- Comparing constant terms: $2 = A + C \implies 2 = 1 + C \implies C = 1$.$$\int \left( \frac{1}{1 – x} + \frac{x + 1}{x^2 + 1} \right) dx = \int \frac{1}{1 – x} dx + \int \frac{x}{x^2 + 1} dx + \int \frac{1}{x^2 + 1} dx$$
- Term 1: $-\log|1 – x|$.
- Term 2: $\frac{1}{2} \log|x^2 + 1|$.
- Term 3: $\tan^{-1} x$.$$= \mathbf{-\log|1 – x| + \frac{1}{2} \log(x^2 + 1) + \tan^{-1} x + C}$$
14. $\int \frac{3x – 1}{(x + 2)^2} dx$
Decomposition form (repeated factor):
$$\frac{3x – 1}{(x + 2)^2} = \frac{A}{x + 2} + \frac{B}{(x + 2)^2}$$
$$3x – 1 = A(x + 2) + B$$
- Put $x = -2$: $3(-2) – 1 = B \implies B = -7$.
- Compare coefficient of $x$: $3 = A \implies A = 3$.$$\int \left( \frac{3}{x + 2} – \frac{7}{(x + 2)^2} \right) dx$$$$= 3 \log|x + 2| – 7 \left(-\frac{1}{x + 2}\right) + C$$$$= \mathbf{3 \log|x + 2| + \frac{7}{x + 2} + C}$$
15. $\int \frac{1}{x^4 – 1} dx = \int \frac{1}{(x^2 – 1)(x^2 + 1)} dx = \int \frac{1}{(x – 1)(x + 1)(x^2 + 1)} dx$
Decomposition form (linear and quadratic factors):
$$\frac{1}{(x – 1)(x + 1)(x^2 + 1)} = \frac{A}{x – 1} + \frac{B}{x + 1} + \frac{Cx + D}{x^2 + 1}$$
$$1 = A(x + 1)(x^2 + 1) + B(x – 1)(x^2 + 1) + (Cx + D)(x^2 – 1)$$
- Put $x = 1$: $1 = A(2)(2) \implies A = 1/4$.
- Put $x = -1$: $1 = B(-2)(2) \implies B = -1/4$.
- Put $x = 0$: $1 = A(1) + B(-1) + D(-1) \implies 1 = 1/4 – (-1/4) – D \implies 1 = 1/2 – D \implies D = -1/2$.
- Compare $x^3$: $0 = A + B + C \implies 0 = 1/4 – 1/4 + C \implies C = 0$.$$\int \left( \frac{1/4}{x – 1} – \frac{1/4}{x + 1} – \frac{1/2}{x^2 + 1} \right) dx$$$$= \frac{1}{4} \log|x – 1| – \frac{1}{4} \log|x + 1| – \frac{1}{2} \tan^{-1} x + C$$$$= \mathbf{\frac{1}{4} \log \left| \frac{x – 1}{x + 1} \right| – \frac{1}{2} \tan^{-1} x + C}$$
Part 4: Integration by Substitution followed by Partial Fractions (16-21)
16. $\int \frac{1}{x(x^n + 1)} dx$
Hint: Multiply numerator and denominator by $x^{n – 1}$ and put $x^n = t$.
$$\int \frac{x^{n – 1}}{x^n(x^n + 1)} dx$$
Let $t = x^n$. Then $dt = n x^{n – 1} dx$, so $x^{n – 1} dx = \frac{1}{n} dt$.
$$\frac{1}{n} \int \frac{1}{t(t + 1)} dt$$
Partial fraction decomposition: $\frac{1}{t(t + 1)} = \frac{1}{t} – \frac{1}{t + 1}$.
$$\frac{1}{n} \int \left( \frac{1}{t} – \frac{1}{t + 1} \right) dt = \frac{1}{n} (\log|t| – \log|t + 1|) + C$$
$$= \mathbf{\frac{1}{n} \log \left| \frac{x^n}{x^n + 1} \right| + C}$$
17. $\int \frac{\cos x}{(1 – \sin x)(2 – \sin x)} dx$
Hint: Put $\sin x = t$.
Let $t = \sin x$. Then $dt = \cos x dx$.
$$\int \frac{dt}{(1 – t)(2 – t)}$$
Partial fraction decomposition: $\frac{1}{(1 – t)(2 – t)} = \frac{A}{1 – t} + \frac{B}{2 – t}$.
$$1 = A(2 – t) + B(1 – t)$$
- Put $t = 1$: $1 = A(1) \implies A = 1$.
- Put $t = 2$: $1 = B(-1) \implies B = -1$.$$\int \left( \frac{1}{1 – t} – \frac{1}{2 – t} \right) dt = \frac{\log|1 – t|}{-1} – \frac{\log|2 – t|}{-1} + C$$$$= \log|2 – t| – \log|1 – t| + C = \log \left| \frac{2 – \sin x}{1 – \sin x} \right| + C$$$$= \mathbf{\log \left| \frac{2 – \sin x}{1 – \sin x} \right| + C}$$
18. $\int \frac{(x^2 + 1)(x^2 + 2)}{(x^2 + 3)(x^2 + 4)} dx$
This is a rational function of $x^2$. First, use long division (or substitution $y = x^2$).
$$\frac{(y + 1)(y + 2)}{(y + 3)(y + 4)} = \frac{y^2 + 3y + 2}{y^2 + 7y + 12}$$
$$= 1 – \frac{4y + 10}{y^2 + 7y + 12} = 1 – \frac{4y + 10}{(y + 3)(y + 4)}$$
Partial fraction decomposition for $\frac{4y + 10}{(y + 3)(y + 4)} = \frac{A}{y + 3} + \frac{B}{y + 4}$.
$$4y + 10 = A(y + 4) + B(y + 3)$$
- Put $y = -3$: $-12 + 10 = A(1) \implies A = -2$.
- Put $y = -4$: $-16 + 10 = B(-1) \implies -6 = -B \implies B = 6$.Integral becomes: $\int \left( 1 – \left( \frac{-2}{x^2 + 3} + \frac{6}{x^2 + 4} \right) \right) dx$$$\int \left( 1 + \frac{2}{x^2 + (\sqrt{3})^2} – \frac{6}{x^2 + 2^2} \right) dx$$$$= x + 2 \cdot \frac{1}{\sqrt{3}} \tan^{-1}\left(\frac{x}{\sqrt{3}}\right) – 6 \cdot \frac{1}{2} \tan^{-1}\left(\frac{x}{2}\right) + C$$$$= \mathbf{x + \frac{2}{\sqrt{3}} \tan^{-1}\left(\frac{x}{\sqrt{3}}\right) – 3 \tan^{-1}\left(\frac{x}{2}\right) + C}$$
19. $\int \frac{2x}{(x^2 + 1)(x^2 + 3)} dx$
Let $t = x^2$. Then $dt = 2x dx$.
$$\int \frac{dt}{(t + 1)(t + 3)}$$
Partial fraction decomposition: $\frac{1}{(t + 1)(t + 3)} = \frac{A}{t + 1} + \frac{B}{t + 3}$.
$$1 = A(t + 3) + B(t + 1)$$
- Put $t = -1$: $1 = A(2) \implies A = 1/2$.
- Put $t = -3$: $1 = B(-2) \implies B = -1/2$.$$\frac{1}{2} \int \left( \frac{1}{t + 1} – \frac{1}{t + 3} \right) dt = \frac{1}{2} (\log|t + 1| – \log|t + 3|) + C$$$$= \mathbf{\frac{1}{2} \log \left| \frac{x^2 + 1}{x^2 + 3} \right| + C}$$
20. $\int \frac{1}{x(x^4 – 1)} dx$
Similar to Q16. Multiply numerator and denominator by $x^3$.
$$\int \frac{x^3}{x^4(x^4 – 1)} dx$$
Let $t = x^4$. Then $dt = 4x^3 dx$, so $x^3 dx = \frac{1}{4} dt$.
$$\frac{1}{4} \int \frac{dt}{t(t – 1)}$$
Partial fraction decomposition: $\frac{1}{t(t – 1)} = \frac{1}{t – 1} – \frac{1}{t}$.
$$\frac{1}{4} \int \left( \frac{1}{t – 1} – \frac{1}{t} \right) dt = \frac{1}{4} (\log|t – 1| – \log|t|) + C$$
$$= \mathbf{\frac{1}{4} \log \left| \frac{x^4 – 1}{x^4} \right| + C}$$
21. $\int \frac{1}{e^x – 1} dx$
Hint: Put $e^x = t$.
Let $t = e^x$. Then $dt = e^x dx = t dx$, so $dx = \frac{dt}{t}$.
$$\int \frac{1}{(t – 1)} \frac{dt}{t} = \int \frac{1}{t(t – 1)} dt$$
Partial fraction decomposition: $\frac{1}{t(t – 1)} = \frac{1}{t – 1} – \frac{1}{t}$.
$$\int \left( \frac{1}{t – 1} – \frac{1}{t} \right) dt = \log|t – 1| – \log|t| + C$$
$$= \mathbf{\log \left| \frac{e^x – 1}{e^x} \right| + C}$$
Part 5: Multiple Choice Questions
22. $\int \frac{x}{(x – 1)(x – 2)} dx$
Partial fraction decomposition (from Q1):
$$\frac{x}{(x – 1)(x – 2)} = \frac{A}{x – 1} + \frac{B}{x – 2}$$
$$x = A(x – 2) + B(x – 1)$$
- Put $x = 1$: $1 = A(-1) \implies A = -1$.
- Put $x = 2$: $2 = B(1) \implies B = 2$.$$\int \left( \frac{-1}{x – 1} + \frac{2}{x – 2} \right) dx = -\log|x – 1| + 2 \log|x – 2| + C = \log \left| \frac{(x – 2)^2}{x – 1} \right| + C$$However, the options are in a different format. Checking option (B):$$\log \left| \frac{(x – 2)^2}{x – 1} \right| + C = 2 \log|x – 2| – \log|x – 1| + C$$This matches the result.
The correct answer is (B) $\log \left| \frac{(x – 2)^2}{x – 1} \right| + C$ (using the form given in the option, although the integral result matches $2 \log|x – 2| – \log|x – 1| + C$).
23. $\int \frac{dx}{x(x^2 + 1)}$
Partial fraction decomposition:
$$\frac{1}{x(x^2 + 1)} = \frac{A}{x} + \frac{Bx + C}{x^2 + 1}$$
$$1 = A(x^2 + 1) + (Bx + C)x$$
- Put $x = 0$: $1 = A(1) \implies A = 1$.
- Compare $x^2$: $0 = A + B \implies B = -1$.
- Compare $x$: $0 = C$.$$\int \left( \frac{1}{x} – \frac{x}{x^2 + 1} \right) dx = \log|x| – \frac{1}{2} \log|x^2 + 1| + C$$$$= \log|x| – \frac{1}{2} \log(x^2 + 1) + C$$This matches option (A) $\log|x| – \frac{1}{2} \log(x^2 + 1) + C$.