Class 12 NCERT Solutions- Mathematics Part ii – Chapter 7– Integrals Exercise 7.5

Various formulas used in the exercise are,

Question 1: x/(x+1)(x+2)

Answer:

Let x/(x+1)(x+2) = A/ (x+1) + B/(x+2)

=> x = A(x+2)+B(x+1)

Equating the coefficient of x and constant term, we obtain

A + B = 1

2A + B = 0

On solving we obtain

A = -1 nd B = 2

So x/(x+1)(x+2) = -1/ (x+1) + 2/(x+2)

=> ∫x/ (x+1)(x+2) dx

= ∫ -1/ (x+1) + 2/(x+2) dx

= -log|x+1| + 2log|x+2| + C

= log(x+2)² – log|x+1| + C

= log (x+2)²/ (x+1) + C

Question 2: 1/(x-9)

Answer:

Let 1/ (x+3)(x-3) = A/ (x+3) + B/(x-3)

1 = A(x-3) + (x+2)

Equating the coefficient of x and constant term, we obtain

A + B = 0

-3A + 3B=1

On solving we obtain

A = -1/6 and B = 1/6

So, 1/ (x+3)(x-3) = -1/6(x+3) + 1/6(x-3)

=> ∫ 1/ (x²-9) dx

= ∫ { -1/6(x+3) + 1/6(x-3)} dx

= -1/6 log| x+3| + 1/6 log|x-3| + C

= 1/6 log| (x-3)/(x+3) | + c

Question 3: (3x-1)/(x-1)(x-2)(x-3)

Answer:

Let (3x-1)/(x-1)(x-2)(x-3) = A/(x-1) + B/ (x-2) + C/(x-3)

3x-1 = A(x-2)(x-3)+B(x-1)(x-3)+C(x-1)(x-2)…(i)

Substituting x = 1, 2 and 3 respectively in equation (i), we obtain

A = 1, B = -5 and C = 4

(3x-1)/(x-1)(x-2)(x-3)

= 1/(x-1) – 5/ (x-2) + 4/(x-3)

=> ∫ (3x-1)/(x-1)(x-2)(x-3) dx

= ∫{1/(x-1) – 5/(x-2) + 4/(x-3) }dx

= log| x-1| – 5log|x-2| +4 log|x-3| + C

Question 4: x/(x-1)(x-2)(x-3)

Answer:

Let x/(x-1)(x-2)(x-3) = A/(x-1) + B/ (x-2) + C/(x-3)

x= A(x-2)(x-3) + B(x-1)(x-3) + C(x-1)(x-2)…(i)

Substituting x = 1, 2 and 3 respectively in equation (i), we obtain

A = 1/2 , B = -2 and C = 3/2

So, x/(x-1)(x-2)(x-3) = 1 /2 (x-1) -2/ (x-2) + 3/2(x-3)

=> ∫x/(x-1)(x-2)(x-3) dx

= ∫ {1 /2 (x-1) -2/ (x-2) + 3/2(x-3) }dx

= 1/2 log |x-1| -2 log|x-2| + 3/2 log|x-3| +C

Question 5: 2x/ (x²+3x+2)

Answer:

Let 2x/ (x²+3x+2) = A/(x+1) + B/(x+2)

2x = A(x+2) + B(x+1)…(i)

Substituting x = -1 and -2 in equation (i), we obtain

A = -2 and B = 4

So,2x/ (x²+3x+2) = -2/(x+1) + 4/(x+2)

=> ∫ 2x/ (x²+3x+2) dx

= ∫ { 4/(x+1) – 2/(x+1)} dx

= 4log|x+2| – 2log|x+1| + C

Question 6: (1-x²)/{x(1-2x)}

Answer:

It can seen that the given integrand is not proper fraction.

Therefore, on diving (1-x²) by x(1-2x), we obtain

(1-x²)/{x(1-2x)} = 1/2 + 1/2 { (2-x)/x(1-2x)}

Let (2-x)/x(1-2x) = A/x B/(1-2x)

=> (2-x) = A(1-2x) + BX…(i)

Substituting x = 0 and 1/2 in equation (i), we obtain

A = 2 and B = 3

So, (2-x)/x(1-2x) = 2/x+ 3/(1-2x)

Substituting in equation (1), we obtain

(1-x²)/{x(1-2x)} = 1/2 + 1/2 { 2/x+ 3/(1-2x)}

=> ∫ (1-x²)/{x(1-2x)} dx

= ∫{ 1/2 + 1/2 { 2/x+ 3/(1-2x)}}dx

= x/2 + log|x| + 3/(2(-2) log|1-2x| + C

= x/2 + log|x| – 3/2 log|1-2x| + C

Question 7: x/(x²+1)(x-1)

Answer:

Let x/(x²+1)(x-1) = (Ax+B)/(x²+1) + C/(x-1)

x = (Ax+B)(x-1) + C(x²+1)

x = Ax² – Ax + Bx-B+Cx² + C

Equating the coefficient of x²,x and constant term, we obtain

A + C = 0

-A + B = 1

-B + C = 0

On solving these equation, we obtain

A = -1/2 , B = 1/2 and C = 1/2

From equation (1), we obtain

So, x/(x²+1)(x-1) = { (-1/2 x + 1/2)/(x²+1)} + (1/2)/(x-1)

=> ∫x/(x²+1)(x-1) = -1/2 ∫x/(x²+1) dx + 1/2 ∫1/(x²+1) dx + 1/2 ∫ 1/ (x-1) dx

=1/4 ∫ 2x/(x²+1) dx + 1/2 tan^-1x + 1/2 log| x-1 | + C

Consider ∫ 2x/(x²+1) dx , let (x²+1) = t => 2x dx = dt

=> ∫ 2x/(x²+1) dx = ∫dt/t = log|t| = log | x²+1|

∫ x/(x²+1)(x-1) dx

= -1/4 log |x²+1| + 1/2 tan^-1x + 1/2 log |x-1| + C

= 1/2 log |x-1| -1/4 log |x²+1| + 1/2 tan^-1x + C

Question 8: x/(x-1)²(x+2)

Answer:

x/(x-1)²(x+2) = A/(x-1) + B/(x-1)² + C/(x+2)

x = A(x-1)(x+2) + B(x+2) + C(x-1)²

Substituting x = 1, we obtain

B = 1/3

Equating the coefficient of x² and constant term, we obtain

A + C = 0

-2A + 2B + C = 0

On solving, we obtain

A = 2/9 and C = -2/9

x/(x-1)²(x+2) = 2 /9(x-1) + 1/3(x-1)2 – 2 /9(x+2)

=> ∫x/(x-1)²(x+2) dx

= 2/9∫1/(x-1) + 1/3 ∫1/(x-1)² dx -2/9 ∫ 1/(x+2) dx

= 2/9 log|x-1| + 1/3 {-1/(x-1)} – 2/9 log |x+2| + C

= 2/9 log |(x-1)/(x+2) | – 1/ 3(x-1) + C

Question 9: (3x+5)/(x³-x²-x+1)

Answer:

Let (3x+5)/(x³-x²-x+1) = A/(x-1) + B/(x-1)² + C/(x+1)

3x + 5 = A(x-1)(x+1) + B(x+1) + C(x-1)²

3x + 5 = A(x²-1) + B(x+1) + C(x²+1-2x)…(i)

Substituting x = 1 in equation (1), we obtain

B = 4

Equating the coefficient of x² and x , we obtain

A+C = 0

B-2C = 3

On solving, we obtain

A = -1/2 and C = 1/2

So, (3x+5)/(x³-x²-x+1) = -1/2(x-1) + 4/(x-1)² + 1/2(x+1)

=> ∫ (3x+5)/{(x-1)²(x+1)} dx

= -1/2∫1/(x-1) dx + 4∫1/(x-1)² dx +1/2 ∫1/(x+1) dx

= -1/2 log | x-1| + 4 {-1/(x-1)} + 1/2 log | x+1 | + C

= 1/2log| (x+1)/ (x-1)| – 4/(x-1) + C

Question 10: (2x-3)/(x²-1)(2x+3)

Answer:

(2x-3)/(x²-1)(2x+3) = (2x-3)/(x+1)(x-1)(2x+3)

Let (2x-3)/(x+1)(x-1)(2x+3) = A/(x+1) + B/(x-1) + C/(2x+3)

=> (2x-3) = A(x-1)(2x+3) + B(2x²+5x+3) + C(x²-1)

=> (2x-3) = A(2x²+x-3) + B(2x²+5x+3) + C(x²-1)

=>(2x-3) = (2A+2B+C)x² + (A+5B)x+(-3A+3B-C)

Equating the coefficient of x² and x , we obtain

B = -1/10 , A = 5/2 and C = -24/5

So, (2x-3)/(x+1)(x-1)(2x+3) = 5/2(x+1) + -1 /10 (x-1) – 24/5 (2x+3)

=> ∫(2x-3)/(x²-1)(2x+3) = 5/2∫1/(x+1) dx – -/10 ∫1/(x-1) dx – 24/5 ∫1/(2x+3) dx

= 5/2 log | x+1| – 1/10 log | x-1 | – 24/5*2 log | 2x+3 | + C

= 5/2 log | x+1| – 1/10 log | x-1 | – 12/5 log | 2x+3 | + C

Question 11: 5x/(x+1)(x²-4)

Answer:

5x/(x+1)(x²-4) = 5x/(x+1)(x+2)(x-2)

Let 5x/(x+1)(x+2)(x-2) = A/(x+1) + B/(x+2) + C/(x-2)

5x = A(x+2)(x-2) + B(x+1)(x-2) + C(x+1)(x+2)…(i)

Substituting x = -1,-2 and 2 in equation (1), we obtain

A = 5/2, B = -5/2 and C = 5/6

So, 5x/(x+1)(x+2)(x-2) = 5 /3(x+1) – 5/2(x+2) + 5 /6 (x-2)

=> ∫ 5x/(x+1)(x+2)(x-2) dx = ∫ 5 /3(x+1) dx – ∫5/2(x+2) dx + ∫ 5 /6 (x-2) dx

= 5/3 log|x+1| -5.2 log|x+2| + 5/6 log |x-2| + C

Question 12: (x³+x+1)/(x²-1)

Answer:

It can seen that the given integrand is not proper fraction.

Therefore, on diving (x³+x+1) by (x²-1), we obtain

(x³+x+1)/(x²-1) = x + (2x+1)/(x²-1)

Let (2x+1)/(x²-1) = A/(x+1) + B/(x-1)

2x+1 = A(x-1) + B(x+1)…(i)

Substituting x = 1,-1 in equation (1), we obtain

A = 1/2 and B = 3/2

So,(x³+x+1)/(x²-1) = x + 1/2(x+1) + 3/2(x-1)

=> ∫ ,(x³+x+1)/(x²-1) dx = ∫x dx + 1/2 ∫1/(x+1) dx + 3/2 ∫ 1/ (x-1) dx

= x² /2 + 1/2 log |x+1| + 3/2 log|x-1| + C

Question 13: 2/(1-x)(1+x²)

Answer:

Let 2/(1-x)(1+x²) = A/(1-x) + (Bx+c)/(1+x²)

2 = A(1+x) + (Bx+C)(1-x)

2 = A+Ax²+Bx-Bx²+C-Cx

Equating the coefficient of x²,x and constant term, we obtain

A-B = 0

B-C = 0

A+C = 2

On solving these equation, we obtain

A = 1, B = 1 and C = 1

So, 2/ (1-x)(1+x²) = 1/(1-x) + (x+1)/(1+x²)

=> ∫2/ (1-x)(1+x²) dx

= ∫1/(1-x) dx + ∫x/(1+x²) dx + ∫ 1/ (1+x²) dx

= – ∫1/(x-1) dx + 1/2∫2x/(1+x²) dx + ∫ 1/ (1+x²) dx

= -log|x-1 | + 1/2 log|1+x²| + tan^-1x + C

Question 14: (3x-1)/(x+2)²

Answer:

Let (3x-1)/(x+2)² = A /(x+2) + B/(x+2)²

=> 3x – 1 = A(x+2) + B

Equating the coefficient of x and constant term, we obtain

A = 3

2A +B = -1 => B = -7

So, (3x-1)/(x+2)² = 3/(x+2) – 7/(x+2)²

=> ∫(3x-1)/(x+2)² dx = 3∫1/(x+2) dx – 7 ∫ x/(x+2)²

= 3log|x+2| -7{-1/(x+2)} +C

= 3log|x+2| + 7/(x+2) + C

Question 15: 1/(x⁴-1)

Answer:

1/(x⁴-1) = 1/(x²-1)(x²+1) = 1/{(x+1)(x-1)(1+x²)}

Let 1/(x⁴-1) = 1/(x²-1)(x²+1) = A /(x+1) + B/(x-1) + (Cx+D)/(1+x²)

1 = A(x-1)(x²+1) + B(x+1)(x²+1) + (Cx+D)(x²-1)

1 = A(x³+x-x²-1)+B(x³+x+x²+1) + Cx³+Dx³-Cx-D

1 = (A+B+C)x³ + (-A+B+D)x² + (A+B-C)x+(-A+B-D)

Equating the coefficient of x³,x²,x and constant term, we obtain

A+B+C = 0

-A+B+D = 0

A+B-C = 0

-A+B-D = 1

On solving these equation, we obtain

a = -1/4, b = 1/4, C = 0 and D = -1/2

∫1/(x⁴-1)dx = ∫-1/4(x+1)dx + ∫1/4(x-1)dx -∫1/2(1+x²)dx

=> ∫1/(x⁴-1)dx = -1/4 log|x-1| + 1/4 log |x–1| -1/2tan^-1x + C

= 1/4 log |(x-1)/(x+1)| – 1/2 tan^-1x + C

Question 16: 1/x(xⁿ+1)

Answer:

1/x(xⁿ+1)

Multiplying numerator and denominator by x^n-1, we obtain

1/x(xⁿ+1) = x^n-1/{x^n-1 x(xⁿ+1)} = x^n-1/xⁿ{xⁿ+1}

Let xⁿ = t => x^n-1 dx = dt

So, ∫1/{x(xⁿ+1)} dx = ∫x^n-1/{xⁿ(xⁿ+1} dx = 1/n ∫1/t(t+1) dt

Let 1 /t(t+1) = A/t + B/(t+1)

1 = A(1+t) + Bt…(i)

Substituting t=0,-1 in equation (i),we obtain

A = 1 and B = -1

So, 1/ t(t+1) = 1/t – 1/(1+t)

=> ∫1/x(xⁿ+1) dx = 1/n ∫ {1/t – 1/(t+1)} dx

= 1/n[log|t| = log|t+1|] + C

= -1/n [ log|xⁿ| – log|xⁿ+1|] +C

= 1/n log |xⁿ/xⁿ+1| + C

Question 17: cosx/(1-sinx)(2-sinx) [ hint: put sinx = t ]

Answer:

cosx/(1-sinx)(2-sinx)

Let sinx = t => cosx dx = dt

So, ∫cosx/(1-sinx)(2-sinx) dx = ∫dt/(1-t)(2-t)

Let 1/(1-t)(2-t) = A/(1-t) + B/(2-t)

1 = A(2-t)+B(1-t)…(i)

Substituting t=2 and then t=1 in equation (i),we obtain

A = 1 and B = -1

So, 1/(1-t)(2-t) = 1/(1-t) – 1/(2-t)

=> ∫cosx/(1-sinx()2-sinx) dx

= ∫ {1/(1-t) – 1/(2-t) }dt

= -log|1-t| + log|2-t|+ C

= log|(2-t)/(1-t)| + C

= log |(2-sinx)/(1-sinx)| + C

Question 18: (x²+1)(x²+2)/(x²+3)(x²+4)

Answer:

(x²+1)(x²+2)/(x²+3)(x²+4) = 1 – (4x²+10)/(x²+3)(x²+4)

Let (4x²+10)/(x²+3)(x²+4) = (Ax+B)/(x²+3) + (Cx+D)/(x²+4)

4x² + 10 = (Ax+B)(x²+4) + (Cx+D)(x²+3)

4x² + 10 = Ax³ + 4Ax + Bx² + 4B + Cx³ + 3Cx + Dx² + 3D

4x² + 10 = (A+C)x³ + (B+D)x² + (4A+3C)x + (4B+3D)

Equating the coefficient of x³,x²,x and constant term, we obtain

A + C = 0

B + D = 4

4A + 3C = 0

4B + 3D = 10

On solving these equation, we obtain

A = 0, B = -2 , C= 0 and D = 6

So, (4x²+10)/(x²+3)(x²+4) = -2/(x²+3) + 6/(x²+4)

(x²+1)(x²+2)/(x²+3)(x²+4) = 1 – { -2/(x²+3) + 6/(x²+4) }

=> ∫(x²+1)(x²+2)/(x²+3)(x²+4) dx

= ∫ 1 + 2/(x²+3) – 6/(x²+4) } dx

= ∫ { 1 + 2/{x²+√(3)² – 6/{x²+2²} } dx

= x + 2{1/√3 tan^-1x/√3} -6{1/2 tan^-1x/2} + C

= x + 2/√3 tan^-1x/√3 – 3 tan^-1x/2 + C

Question 19: 2x/(x²+1)(x³+3)

Answer:

2x/(x²+1)(x³+3)

Let x² = t => 2x dx = dt

∫ 2x/(x²+1)(x²+3) dx = ∫dt/(t+1)(t+3)…(i)

Let 1/(t+1)(t+3) = A/(t+1) + B?(t+3)

1 = A(t+3) + B(t+1)…(ii)

Substituting t = -3 and then t = -1 in equation (i), we obtain

A = 1/2 and B = -1/2

So, 1/(t+1)(t+3) = 1/2(t+1) – 1/2(t+3)

=>∫2x/(x²+1)(x²+3) dx = ∫{1/2(t+1) – 1/2(t+3)} dt

1/2 log| (t+1) |-1/2 log|t+3| + C

= 1/2 log |(t+1)/(t+3)| + C

= 1/2 log |(x²+1)/(x²+3)| + C

Question 20: 1/x(x⁴-4)

Answer:

1/x(x⁴-1)

Multiplying numerator and denominator by x³, we obtain

1/x(x⁴-1) = x³/ x⁴(x⁴-1)

∫ 1/x(x⁴-1) dx = ∫x ³/ x⁴(x⁴-1) dx

Let x⁴ = t => 4x³ dx = dt

∫ 1/x(x⁴-1) dx =1/4 ∫ dt/ t(t-1)

Let 1/t(t-1) = A/t + B/(t-1)

1 = A(t-1) + Bt…(i)

Substituting t= 0 and 1 in equation (i),we obtain

A = -1 and B = 1

=> ∫ 1/x(x⁴-1) dx = 1/4 ∫ { -1/t + 1/ (t-1) } dt

= 1/4 [-log|t| + log |t-1 | ] + C

= 1/4 log |(t-1)/t| + C

= 1/4 log | (x⁴-1)/x⁴ | + C

Question 21: 1/(e^x-1) [ Hint: put e^x = t]

Answer:

1/(e^x-1)

Let e^x = t => e^x dx = dt

=> ∫1/ (e^x-1) dx = ∫1/(t-1) dt/t = ∫ 1/t(t-1) dt

Let 1/t(t-1) = A/t + B/(t-1)

1 = A(t-1) + Bt…(i)

Substituting t = 1 and t= 0 in equation (i), we obtain

A = -1 and B = 1

So, 1/t(t-1) = -1/t + 1/(t-1)

=> ∫ 1/t(t-1) dt

= log | (t-1)/t | + C

= log | (e^x-1)/ex | + C

Question 22: ∫ x dx/(x-1)(x-2) equals

(A) log | (x-1)²/(x-2) | + C

(B) log | (x-2)²/(x-1) | + C

(C) log | {(x-1)/(x-2)}² | + C

(D) log | (x-1)(x-2) | + C

Correct Answer is option (B)

Question 23: ∫ dx/x(x²+1) equals

(A) log|x| – 1/2 log(x²+1) + C

(B) log|x| + 1/2 log(x²+1) + C

(C) – log|x| + 1/2 log(x²+1) + C

(D) 1/2 log|x| + log(x²+1) + C

Correct Answer is option (A)