# 2nd PUC Basic Maths Question Bank Chapter 20 Indefinite Integrals Ex 20.5

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## Karnataka 2nd PUC Basic Maths Question Bank Chapter 20 Indefinite Integrals Ex 20.5

Part-A

Integration by partial fractions

2nd PUC Basic Maths Indefinite Integrals Ex 20.5 Five Marks Questions and Answers

Question 1.
$$\int \frac{4 x+5}{(x-1)(x+2)} d x$$
Let

4x + 5 = A(x + 2) + B(x – 1)
put x = 1, 4 + 5 = A(1 + 2) + B(0)
9 = 3A ⇒ A = 3

put x = -2 -8 + 5 = A(0) + B(-2 -1)
-3 = -3B ⇒ B = 1

∴$$\int \frac{4 x+5}{(x-1)(x+2)} d x=\int\left(\frac{3}{x-1}+\frac{1}{x+2}\right) d x$$ =3log(x – 1) + log(x + 2) +c

Question 2.
$$\int \frac{3 x+2}{(2 x+3)(3 x-1)} d x=\int \frac{A}{2 x+3}+\frac{B}{3 x-1} \cdot d x$$
$$\int \frac{3 x+2}{(2 x+3)(3 x-1)} d x=\int \frac{A}{2 x+3}+\frac{B}{3 x-1} \cdot d x$$
Let 3x + 2 = A(3x – 1) + B(2x + 3)
comparing coefficients of x both sides & Constant terms

Question 3.
$$\int \frac{1}{x(x+1)(x+2)} d x$$
Let
$$\frac{1}{x(x+1)(x+2)}=\frac{A}{x}+\frac{B}{x+1}+\frac{C}{x+2}$$ …..(1)
1 = A ( x + 1) (x + 2) +B(x) (x + 2) + c × x(x + 1)
put x = 0, 1 = A(1)(2) ⇒ 1 = 2A ⇒ A = $$\frac { 1 }{ 2 }$$
put x = -1, 1 = A(0) + B(-1) (-1 + 2) + c(0) ⇒ 1 = -B ⇒ B = -1
put x = -2 1 =A(0) + B(0) + c(-2), (-2 + 1) ⇒ 1 = 2c ⇒ c = $$\frac { 1 }{ 2 }$$
∴$$\int \frac{d x}{x(x+1)(x+2)}=\int \frac{\frac{1}{2}}{x}+\frac{-1}{x+1}+\frac{\frac{1}{2}}{x+2} \cdot d x$$
= $$\frac { 1 }{ 2 }$$log 2 – log(x + 1) + $$\frac { 1 }{ 2 }$$log(x + 2) +c

Question 4.
$$\int \frac{5 x+7}{(x-2)^{2}(x+3)} d x$$
Let
$$\frac{5 x+7}{(x-2)^{2}(x+3)}=\frac{A}{x-2}+\frac{B}{(x-2)^{2}}+\frac{C}{x+3}$$
5x + 7 = A(x – 2) (x + 3) + B(x +3) + (x – 2)2
put x = 2 10 + 7 = A(0) + B(2 + 3) + c(0)2 = 17 = 5B ⇒ B = $$\frac { 17 }{ 5 }$$
put x = -3 -15 + 7 = A(0) + B(0) + c(-5)2 = -8 = 25c ⇒ c = $$\frac { 17 }{ 5 }$$
Comparing the coefficient of ×2 both sides we get 0 = A + c ⇒ A = -c = $$\frac { 8 }{ 25 }$$

Question 5.
$$\int \frac{5}{\left(x^{2}+6 x+9\right)(x-3)} d x$$
$$\int \frac{5}{\left(x^{2}+6 x+9\right)(x-3)} d x=\int \frac{5 d x}{(x+3)^{2}(x-3)}=\int \frac{A}{x+3}+\frac{B}{(x+3)^{2}}+\frac{C}{(x-3)} d x$$
Let
$$\frac{5}{(x+3)^{2}(x-3)}=\frac{A}{x+3}+\frac{B}{(x+3)^{2}}+\frac{C}{x-3}$$ ….(1)
5 = A(x + 3) (x – 3) + B(x – 3) + c(x + 3)2
put x = 3, 5 = A(0) + B(0) + C(3 +3)2 = 5 = 36C ⇒ c = $$\frac { 5 }{ 35 }$$
put x = -3, 5 = (A) (0) + B(-3-3) + C(0)2
5 = -6B ⇒C = $$\frac { 5 }{ 36 }$$
put x = -3, 5 = (A) (0) + B(-3-3) + C(0)2
5 = -6B ⇒ B = $$\frac { 5 }{ -6 }$$
Comparing coefficients of ×2 both sides we get
0 = A + C ⇒ A = -C = $$\frac { -5 }{ 36 }$$

Question 6.
$$\int \frac{2 x-1}{\left(x^{2}-4\right)(x+1)} d x$$
Let
$$\int \frac{(2 x-1) d x}{(x-2)(x+2)(x+1)}=\int \frac{A}{x-2}+\frac{B}{x+2}+\frac{C}{x+1} \cdot d x$$ …(1)
2x – 1 = A(x + 2) (x + 1) +B(x – 2) (x + 1) + c(x – 2)(x + 2)
put x = 2 4 – 1 = A(4) (3) + B(0) + C(0)
3 = 12A ⇒ A = $$\frac { 1 }{ 4 }$$
put x = 2 -4 -1 = A(0) + B(-2) (-2 +1) + C(0)
-5 = 4B ⇒ B = $$\frac { -5 }{ 36 }$$
put x = -1 -2 -1 = A(0) + B(0) + C(-3) 1
-3 = -3c ⇒ c = 1

Question 7.
$$\int \frac{3 e^{x}}{e^{2 x}+5 e^{x}+6} d x$$
$$\int \frac{3 e^{x}}{\left(e^{x}\right)^{2}+5 e^{x}+6} d x=3 \int \frac{t}{t^{2}+5 t+6} d t$$ put ex = t
$$=1 \int \frac{3 t d t}{(t+2)(t+3)}=\int \frac{A}{t+2}+\frac{B}{t+3} d t$$
Let 36 = A(t + 3) + B(t + 2)
put t = -2 -6 = A(1) ⇒ A = -6
put t = -3, -9 = A(0) + B(-1) ⇒ B = 9
∴ $$\int \frac{3 e^{x}}{\left(e^{x}\right)^{2}+5 e^{x}+6}=\int \frac{-6}{t+2}+\frac{9}{t+3} \cdot d t$$
= -6 log (t + 2) + 9 log (t + 3) + C = -6 log(ex + 3) + 9 log (ex + 3) + c

Question 8.
$$\int \frac{6}{x(2 \log x)^{2}+7 \log x+5} d x$$
$$\int \frac{6 / x}{(2 \log x)^{2}+7 \log x+5} d x=\int \frac{6 \cdot d t}{2 t^{2}+7 t+5}$$
put log x = t
$$\int \frac{6 d t}{2 t^{2}+7 t+5}=\int \frac{6 d t}{(2 t+5)(t+1)}=\int \frac{A}{2 t+5}+\frac{B}{t+1} \cdot d t$$
Let 6 = A(t + 1) + B(2t + 5)
put t = -1, 6 = B(3) ⇒ B = 2
Equate the coefficient of t both sides
0 = A + 2B
A = -2B = -4
∴ $$\int \frac{6 d x}{x(2 \log x)^{2}+7 \log x+5}=\int \frac{-4}{2 t+5}+\frac{2}{t+1} \cdot d t=\frac{-4 \log (2 t+5)}{2}+2 \log (t+1)$$
= – 2log(2 log x + 5) + 2log(logx + 1) + C

Question 9.
$$\int \frac{3 x^{2}+2 x+3}{(x+1)(3 x+2)} d x$$
$$\int \frac{3 x^{2}+2 x+3}{(x+1)(3 x+2)} d x$$ this is an improper fraction.
$$\int \frac{x^{2}+3 x-2}{x^{2}-4 x-12} d x$$
put x = 6 42 + 10 = A(6 + 2) + B(0); 52 = 8A ⇒ A = $$\frac{B}{2}$$
-4 = -8B ⇒ B = $$\frac{-4}{-8}=\frac{1}{2}$$