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

Students can Download Basic Maths Exercise 20.5 Questions and Answers, Notes Pdf, 2nd PUC Basic Maths Question Bank with Answers helps you to revise the complete Karnataka State Board Syllabus and score more marks in your examinations.

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.
\(\int \frac{4 x+5}{(x-1)(x+2)} d x\)
Answer:
Let
2nd PUC Basic Maths Question Bank Chapter 20 Indefinite Integrals Ex 20.5 - 8
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

KSEEB Solutions

Question 2.
\(\int \frac{3 x+2}{(2 x+3)(3 x-1)} d x=\int \frac{A}{2 x+3}+\frac{B}{3 x-1} \cdot d x\)
Answer:
\(\int \frac{3 x+2}{(2 x+3)(3 x-1)} d x=\int \frac{A}{2 x+3}+\frac{B}{3 x-1} \cdot d x\)
Let 3x + 2 = A(3x – 1) + B(2x + 3)
comparing coefficients of x both sides & Constant terms
2nd PUC Basic Maths Question Bank Chapter 20 Indefinite Integrals Ex 20.5 - 1

Question 3.
\(\int \frac{1}{x(x+1)(x+2)} d x\)
Answer:
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

KSEEB Solutions

Question 4.
\(\int \frac{5 x+7}{(x-2)^{2}(x+3)} d x\)
Answer:
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 = \(\frac { 8 }{ 25 }\)
2nd PUC Basic Maths Question Bank Chapter 20 Indefinite Integrals Ex 20.5 - 2

Question 5.
\(\int \frac{5}{\left(x^{2}+6 x+9\right)(x-3)} d x\)
Answer:
\(\int \frac{5}{\left(x^{2}+6 x+9\right)(x-3)} d x=\int \frac{5 d x}{(x+3)^{2}(x-3)}=\int \frac{A}{x+3}+\frac{B}{(x+3)^{2}}+\frac{C}{(x-3)} d x\)
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 = \(\frac { -5 }{ 36 }\)
2nd PUC Basic Maths Question Bank Chapter 20 Indefinite Integrals Ex 20.5 - 3

KSEEB Solutions

Question 6.
\(\int \frac{2 x-1}{\left(x^{2}-4\right)(x+1)} d x\)
Answer:
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
2nd PUC Basic Maths Question Bank Chapter 20 Indefinite Integrals Ex 20.5 - 4

Question 7.
\(\int \frac{3 e^{x}}{e^{2 x}+5 e^{x}+6} d x\)
Answer:
\(\int \frac{3 e^{x}}{\left(e^{x}\right)^{2}+5 e^{x}+6} d x=3 \int \frac{t}{t^{2}+5 t+6} d t\) 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

KSEEB Solutions

Question 8.
\(\int \frac{6}{x(2 \log x)^{2}+7 \log x+5} d x\)
Answer:
\(\int \frac{6 / x}{(2 \log x)^{2}+7 \log x+5} d x=\int \frac{6 \cdot d t}{2 t^{2}+7 t+5}\)
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.
\(\int \frac{3 x^{2}+2 x+3}{(x+1)(3 x+2)} d x\)
Answer:
\(\int \frac{3 x^{2}+2 x+3}{(x+1)(3 x+2)} d x\) this is an improper fraction.
2nd PUC Basic Maths Question Bank Chapter 20 Indefinite Integrals Ex 20.5 - 5

KSEEB Solutions

Question 10.
\(\int \frac{x^{2}+3 x-2}{x^{2}-4 x-12} d x\)
Answer:
2nd PUC Basic Maths Question Bank Chapter 20 Indefinite Integrals Ex 20.5 - 6
Let 7x + 10 = A(x + 2) + B(x – 6)
put x = 6 42 + 10 = A(6 + 2) + B(0); 52 = 8A ⇒ A = \(\frac{B}{2}\)
put x = -2 -14 + 10 = A(0) + B(-2 -6)
-4 = -8B ⇒ B = \(\frac{-4}{-8}=\frac{1}{2}\)
2nd PUC Basic Maths Question Bank Chapter 20 Indefinite Integrals Ex 20.5 - 7

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