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

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.

\(\int \frac{3 x+2}{(2 x+3)(3 x-1)} d x=\int \frac{A}{2 x+3}+\frac{B}{3 x-1} \cdot d x\)

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

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

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 = \(\frac { 8 }{ 25 }\)

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 = \(\frac { -5 }{ 36 }\)

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

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

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.

Question 10.

\(\int \frac{x^{2}+3 x-2}{x^{2}-4 x-12} d x\)

Answer:

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 = \(\frac{-4}{-8}=\frac{1}{2}\)