Practicing the Intermediate 2nd Year Maths 2B Textbook Solutions Inter 2nd Year Maths 2B Integration Solutions Exercise 6(e) will help students to clear their doubts quickly.

## Intermediate 2nd Year Maths 2B Integration Solutions Exercise 6(e)

I. Evaluate the following integrals.

Question 1.
∫$$\frac{x-1}{(x-2)(x-3)}$$dx
Solution:

Question 2.
∫$$\frac{x^2}{(x+1)(x+2)^2}$$dx
Solution:
∫$$\frac{x^2}{(x+1)(x+2)^2}$$ ≡ $$\frac{A}{x+1}+\frac{B}{x+2}+\frac{C}{(x+2)^2}$$
⇒ x² = A(x + 2)² + B(x + 1)(x + 2) + (x + 1) …………….. (1)
Put x = -2 in (1)
(-2)² = A(0) + B(0) + C(-2 + 1) ⇒ C = -4
Put x = -1 in (1)
(-1)² = A(-1 + 2)² + B(0) + C(0)
⇒ A = 1
Equation coeffs. of x² in (1)
1 = A + B
⇒ B = 1 – A = 1 – 1 = 0
∴ $$\frac{x^2}{(x+1)(x+2)^2}=\frac{1}{x+1}+\frac{0}{x+2}+\frac{(-4)}{(x+2)^2}$$
∴ ∫$$\frac{x^2}{(x+1)(x+2)^2}$$dx
= ∫$$\frac{1}{x+1}$$dx – 4∫$$\frac{1}{(x+2)^2}$$dx
= log|x + 1| – 4$$\frac{(-1)}{x+2}$$
= log|x + 1| + $$\frac{4}{x+2}$$ + C

Question 3.
∫$$\frac{x+3}{(x-1)(x^2+1)}$$dx
Solution:
Let $$\frac{x+3}{(x-1)(x^2+1)}=\frac{A}{x-1}+\frac{Bx+C}{x^2+1}$$
⇒ (x + 3) = A(x² + 1) + (Bx + C)(x – 1) …………….. (1)
Put x = 0 in (1)
3 = A(1) + C(-1)
⇒ A – C = 3 ⇒ C = A – 3 = 2 – 3 = -1
Equation coefficient of x² in (1)
0 = A + B
⇒ B = -A = -2
∴ $$\frac{x+3}{(x-1)(x^2+1)}=\frac{+2}{(x-1)}+\frac{-2x-1}{x^2+1}$$
∫$$\frac{x+3}{(x-1)(x^2+1)}$$dx = 2∫$$\frac{1}{x-1}$$dx
-∫$$\frac{2x}{x^2+1}$$dx – ∫$$\frac{1}{x^2+1}$$dx
= 2 log |x – 1| – log |x² + 1| – tan-1(x) + C

Question 4.

Solution:

Question 5.
∫$$\frac{dx}{(e^x+e^{2x}}$$
Solution:

Question 6.
∫$$\frac{dx}{(x+1)(x+2)}$$
Solution:

Question 7.
∫$$\frac{1}{(e^x-1}$$dx
Solution:

Question 8.
∫$$\frac{1}{(1-x)(4+x^2)}$$dx
Solution:
Let $$\frac{1}{(1-x)(4+x^2)}=\frac{A}{1-x}+\frac{Bx+C}{4+x^2}$$
⇒ 1 = A(4 + x²) + (Bx + C)(1 – x) ………….. (1)
Put x = 1 in (1)
1 = A(4 + 1) ⇒ A = $$\frac{1}{5}$$
Put x = 0 in (1)
1 = A(4) + C(1)
⇒ C = 1 – 4A = 1 – 4($$\frac{1}{5}$$) = $$\frac{5-4}{5}$$ = $$\frac{1}{5}$$
0 = A – B
⇒ B = A = $$\frac{1}{5}$$

Question 9.
∫$$\frac{2x+3}{x^3+x^2-2x}$$dx
Solution:

Put x = 0 in (1), then
3 = A(2)(-1) + B(0) + C(0)
⇒ A = –$$\frac{3}{2}$$
Put x = 1 in (1). Then
2 + 3 = A(0) + B(0) + C(1)(3)
⇒ C = $$\frac{5}{3}$$
Put x = -2 in (1). Then
2(-2) + 3 = A(0) + B(-2)(-2 – 1) + C(0)
⇒ -1 = 6B ⇒ B = $$\frac{-1}{6}$$

II. Evaluate the following integrals.

Question 1.
∫$$\frac{dx}{6x^2-5x+1}$$
Solution:

Question 2.
∫$$\frac{dx}{x(x+1)(x+2)}$$
Solution:
$$\frac{1}{x(x+1)(x+2)}$$ ≡ $$\frac{A}{x}+\frac{B}{x+1}+\frac{C}{x+2}$$
⇒ 1 ≡ A(x + 1)(x + 2) + B(x)(x + 2) + C(x)(x + 1)
Put x = 0
1 = A(1)(2) + B(0) + C(0) ⇒ A = $$\frac{1}{2}$$
Put x = -1
1 = A(0) + B(0) + C(-2)(-2 + 1)
⇒ C = $$\frac{1}{2}$$

Question 3.
∫$$\frac{3x-2}{(x-1)(x+2)(x-3)}$$dx
Solution:
$$\frac{3x-2}{(x-1)(x+2)(x-3)}$$ ≡ $$\frac{A}{x-1}+\frac{B}{x+2}+\frac{C}{x-3}$$
⇒ 3x – 2 = A(x + 2)(x – 3) + B(x – 1)(x – 3) + C(x – 1)(x + 2)
Put x = 1
3(1) – 2 = A(1 + 2)(1 – 3) + B(0) + C(0)
⇒ A = $$\frac{-1}{6}$$
Put x = 3
3(3) – 2 = A(0) + B(0) + C(3 – 1)(3 + 2)
C = $$\frac{7}{10}$$
Put x = -2
3(-2) – 2 = A(0) + B(-2 – 1)(-2 – 3) + C(0) – 8
= 15B ⇒ B = $$\frac{-8}{15}$$

Question 4.
∫$$\frac{7x-4}{(x-1)^2(x+2)}$$dx
Solution:
$$\frac{7x-4}{(x-1)^2(x+2)}$$ ≡ $$\frac{A}{x-1}+\frac{B}{(x-1)^2}+\frac{C}{x+2}$$
⇒ 7x – 4 = A(x – 1)(x + 2) + B(x + 2) + C(x – 1)² ………….. (1)
Put x = 1 in (1)
7 – 4 = A(0) + B(1 + 2) ⇒ B = 1
Put x = -2 in (1)
7(-2) – 4 = A(0) + B(0) + C(-2 – 1)²
⇒ -18 = 9C ⇒ C = -2
Equating coeffs. of x² in (1)
0 = A + C ⇒ A = -C = 2

III. Evaluate the following integrals.

Question 1.
∫$$\frac{1}{(x-a)(x-b)(x-c)}$$dx
Solution:

Question 2.
∫$$\frac{2x+3}{(x+3)(x^2+4)}$$dx
Solution:
$$\frac{2x+3}{(x+3)(x^2+4)}$$ = $$\frac{A}{x+3}+\frac{Bx+C}{x^2+4}$$
2x + 3 = A(x² + 4) + (Bx + C)(x + 3)
x = -3 ⇒ -3 = A(9 + 4) = 13A
A = –$$\frac{3}{13}$$
Equating the coefficient of x²
0 = A + B ⇒ B = -A = $$\frac{3}{13}$$
Equating the constants
3 = 4A + 3C

Question 3.
∫$$\frac{2x^2+x+1}{(x+3)(x-2)^2}$$dx
Solution:
Let $$\frac{2x^2+x+1}{(x+3)(x-2)^2}$$ = $$\frac{A}{x+3}+\frac{B}{x-2}+\frac{C}{(x-2)^2}$$
2x² + x + 1 = A(x – 2)² + B(x + 3)(x – 2) + C(x + 3)
x = 2 ⇒ 8 + 2 + 1 = C(2 + 3) = 5C
⇒ C = $$\frac{11}{5}$$
x = -3 ⇒ 18 – 3 + 1
= A(-5)² = 25 A ⇒ A = $$\frac{16}{25}$$
Equating the coefficients of x²
2 = A + B ⇒ B = 2 – A = 2 – $$\frac{16}{25}$$ = $$\frac{34}{25}$$

Question 4.
∫$$\frac{dx}{x^3+1}$$dx
Solution:
$$\frac{1}{x^3+1}$$ = $$\frac{1}{(x+1)(x^2-x+1)}$$
Let $$\frac{1}{x^3+1}$$ = $$\frac{1}{x+1}+\frac{1}{x^2-x+1}$$
⇒ 1 = A(x² – x + 1) + (Bx + C)(x + 1) ……………. (1)
Put x = -1 in (1)
1 = A(1 + 1 + 1) + (-B + C)(0)
⇒ 3A = 1 ⇒ A = $$\frac{1}{3}$$
Put x = 0 in (1)
1 = A(1) + C(1)
⇒ C = 1 – A = 1 – $$\frac{1}{3}=\frac{2}{3}$$
Equating the coefficients of x²
O = A + B ⇒ B = -A = –$$\frac{1}{3}$$

Question 5.
∫$$\frac{\sin x \cos x}{\cos^2 x+3cos x+2}$$dx
Solution:
Put cos x = t ⇒ – sin x dx = dt
∫$$\frac{\sin x \cos x}{\cos^2 x+3cos x+2}$$dx = ∫$$\frac{-t dt}{t^2+3t+2}$$
= -∫$$\frac{t}{t^2+3t+2}$$dt …………. (1)
Let $$\frac{t}{t^2+3t+2}$$ = $$\frac{t}{(t+1)(t+2)}$$
= $$\frac{A}{t+1}+\frac{B}{t+2}$$
⇒ t = A(t + 2) + B(t + 1) ………… (2)
Put t = -1 in (2)
-1 = A(-1 + 2) ⇒ A = -1
Put t = -2 in (2)
-2 = B(-2 + 1) ⇒ B = 2
∴ $$\frac{t}{t^2+3t+2}$$ = $$\frac{-1}{t+1}+\frac{2}{t+2}$$ ……….. (3)
∴ From (1) & (2)
∫$$\frac{\sin x.\cos x}{\cos^2 x+3cos x+2}$$dx
= -[∫$$\frac{-1}{t+1}$$dt+2∫$$\frac{1}{t+2}$$]
= ∫$$\frac{1}{t+1}$$ – 2∫$$\frac{1}{t+2}$$
= log|t + 1| – 2log|t + 2| + C
= log|1 + cos x| – 2log|2 + cos x| + C
= log|1 + cos x| – log(2 + cos x)² + C
= log|$$\frac{1+\cos x}{(2+\cos x)^2}$$| + C