{"id":31912,"date":"2022-03-29T15:30:23","date_gmt":"2022-03-29T10:00:23","guid":{"rendered":"https:\/\/mcq-questions.com\/?p=31912"},"modified":"2022-03-29T15:41:20","modified_gmt":"2022-03-29T10:11:20","slug":"ncert-solutions-for-class-12-maths-chapter-7-ex-7-5","status":"publish","type":"post","link":"https:\/\/mcq-questions.com\/ncert-solutions-for-class-12-maths-chapter-7-ex-7-5\/","title":{"rendered":"NCERT Solutions for Class 12 Maths Chapter 7 Integrals Ex 7.5"},"content":{"rendered":"

These NCERT Solutions for Class 12 Maths<\/a> Chapter 7 Integrals Ex 7.5 Questions and Answers are prepared by our highly skilled subject experts. https:\/\/mcq-questions.com\/ncert-solutions-for-class-12-maths-chapter-7-ex-7-5\/<\/p>\n

NCERT Solutions for Class 12 Maths Chapter 7 Integrals Exercise 7.5<\/h2>\n

\"NCERT<\/p>\n

Exercise 7.5 Class 12 Maths Solutions Question 1.<\/strong>
\n\\(\\frac { x }{ (x+1)(x+2) }\\)
\nSolution:
\nSolution:
\nlet \\(\\frac { x }{ (x+1)(x+2) }\\) \u2261 \\(\\frac { A }{ x+1 } +\\frac { B }{ x+2 } \\)
\n\u2234 x = A(x + 2) + B(x + 1) … (i)
\nputting x = -1 & x = -2 in (i)
\nwe get A = 1, B = 2
\n\\(\\frac{x}{(x+1)(x+2)}=\\frac{-1}{x+1}+\\frac{2}{x+2}\\)
\nIntegrating w.r.t. x, we get
\n\\(\\int \\frac{x}{(x+1)(x+2)} d x=\\int \\frac{-1}{x+1} d x+\\int \\frac{2}{x+2} d x\\)
\n\\(\\frac{x}{(x+1)(x+2)}=\\frac{-1}{x+1}+\\frac{2}{x+2}\\)
\n= – log|x + 1| + 2log|x + 2| + C
\n= – log|x + 1| + log(x + 2)\u00b2 + C
\n= log\\(\\frac{(x+2)^{2}}{|x+1|}\\) + C<\/p>\n

Ex 7.5 Class 12 NCERT Solutions Question 2.<\/strong>
\n\\(\\frac { 1 }{ { x }^{ 2 }-9 } \\)
\nSolution:
\nLet
\n\\(\\frac { 1 }{ { x }^{ 2 }-9 } =\\frac { 1 }{ (x-3)(x+3) } \\equiv \\frac { A }{ x-3 } +\\frac { B }{ x+3 } \\)
\n\u21d2 1 = A(x + 3) + B(x – 3) … (i)
\nPut x = 3 in (1), we get 1 = 6B \u2234 B = \\(\\frac { 1 }{ 6 }\\)
\nPut x = – 3 in (1), we get 1 = – 6A \u2234 A = \\(\\frac { – 1 }{ 6 }\\)
\n\"NCERT<\/p>\n

Question 3.
\n\\(\\frac { 3x-1 }{ (x-1)(x-2)(x-3) }\\)
\nSolution:
\nLet
\n\\(\\frac { 3x-1 }{ (x-1)(x-2)(x-3) } =\\frac { A }{ x-1 } +\\frac { B }{ x-2 } +\\frac { C }{ x-3 } \\)
\n\u2234 3x – 1 = A(x – 2)(x – 3) + B(x – 1)(x – 3) + C(x – 1)(- 2) … (i)
\nPut x = 1 in (1), we get 2 = 2A \u2234 A = 1
\nPut x = 2 in (1), we get 5 = – B \u2234 B = – 5
\nPut x = 3 in (1), we get 8 = 2C \u2234 C = 4
\n\"NCERT<\/p>\n

\"NCERT<\/p>\n

Question 4.
\n\\(\\frac { x }{ (x-1)(x-2)(x-3) }\\)
\nSolution:
\nlet
\n\\(\\frac { x }{ (x-1)(x-2)(x-3) } =\\frac { A }{ x-1 } +\\frac { B }{ x-2 } +\\frac { C }{ x-3 } \\)
\nx = A(x – 2)(x – 3) + B(x – 1)(x – 3) + C(x – 1)(x – 2) … (i)
\nPut x = 1 in (1), we get 1 = 2A \u2234 A = \\(\\frac { 1 }{ 2 }\\)
\nPut x = 2 in (1), we get 2 = – B \u2234 B = – 2
\nPut x = 3 in (1), we get 3 = 2C \u2234 C = \\(\\frac { 3 }{ 2 }\\)
\n\"NCERT<\/p>\n

Question 5.
\n\\(\\frac { 2x }{ { x }^{ 2 }+3x+2 } \\)
\nSolution:
\nlet
\n\\(\\frac { 2x }{ { x }^{ 2 }+3x+2 } =\\frac { 2x }{ (x+1)(x+2) } =\\frac { A }{ x+1 } +\\frac { B }{ x+2 } \\)
\n\u21d2 2x = A(x + 2) + B(x + 1) … (i)
\nput x = -1, -2 in (i)
\nwe get A = -2, B = 4
\n\\(\\int \\frac{2 x}{(x+1)(x+2)} d x=-2 \\int \\frac{1}{x+1} d x+4 \\int \\frac{1}{x+2} d x\\)
\n= – 2log|x + 1| + 4log|x + 2| + C
\n= 4log|x + 2| – 2log|x + 1| + C<\/p>\n

Question 6.
\n\\(\\frac { 1-{ x }^{ 2 } }{ x(1-2x) } \\)
\nSolution:
\n\"NCERT<\/p>\n

\"NCERT<\/p>\n

Question 7.
\n\\(\\frac { x }{ \\left( { x }^{ 2 }+1 \\right) \\left( x-1 \\right) } \\)
\nSolution:
\nlet
\n\\(\\frac { x }{ \\left( { x }^{ 2 }+1 \\right) \\left( x-1 \\right) } =\\frac { A }{ x-1 } +\\frac { Bx+C }{ { x }^{ 2 }+1 } \\)
\n\u21d2 x = A(x\u00b2 + 1) + (Bx + C)(x – 1) … (1)
\nPut x = 1 in (1), we get 1 = 2A
\n\u2234 A = \\(\\frac { 1 }{ 2 }\\)
\nEquating the coefficients of x\u00b2 and constant term, we get A + B = 0 and A – C = 0
\n\u2234 B = \\(\\frac { 1 }{ 2 }\\) and C = \\(\\frac { 1 }{ 2 }\\)
\n\"NCERT<\/p>\n

Question 8.
\n\\(\\frac { x }{ { \\left( x-1 \\right) }^{ 2 }\\left( x+2 \\right) } \\)
\nSolution:
\n\\(\\frac { x }{ { \\left( x-1 \\right) }^{ 2 }\\left( x+2 \\right) } =\\frac { A }{ x-1 } +\\frac { B }{ { \\left( x-1 \\right) }^{ 2 } } +\\frac { C }{ x+2 } \\)
\n\u21d2 x = A(x – 1)(x + 2) + B(x + 2) + C(x – 1)\u00b2 … (i)
\nPut x = 1 in (1), we get 1 = 3B \u2234 B = \\(\\frac { 1 }{ 3 }\\)
\nPut x = – 2 in (1), we get – 2 = 9C \u2234 C = \\(\\frac { – 2 }{ 9 }\\)
\nEquating the coefficients of x\u00b2, we get A + C = 0
\nHence A = \\(\\frac { 2 }{ 9 }\\)
\n\"NCERT<\/p>\n

Question 9.
\n\\(\\frac { 3x+5 }{ { x }^{ 3 }-{ x }^{ 2 }-x+1 } \\)
\nSolution:
\nx\u00b3 – x\u00b2 – x + 1 = x\u00b2(x – 1) – 1(x – 1)
\n= (x\u00b2 – 1)(x – 1)
\n= (x + 1) (x – 1) (x – 1)
\n= (x + 1)(x – 1)\u00b2
\n\u2234 \\(\\frac { 3x+5 }{ { x }^{ 3 }-{ x }^{ 2 }-x+1 } \\) = \\(\\frac{3 x+5}{(x+1)(x-1)^{2}}\\)
\nLet \\(\\frac{3 x+5}{(x+1)(x-1)^{2}}=\\frac{\\mathrm{A}}{x+1}+\\frac{\\mathrm{B}}{(x-1)}+\\frac{\\mathrm{C}}{(x-1)^{2}}\\)
\n3x + 5 = A(x – 1) + B(x – 1)(x + 1) + C (x + 1) … (1)
\nPut x = – 1 in (1), we get 2 = 4A \u2234 A = \\(\\frac { 1 }{ 2 }\\)
\nPut x = 1 in (1), we get 8 = 2C \u2234 C = 4
\nEquating the coefficients of x\u00b2, we get A + B = 0
\n\u2234 B = \\(\\frac { -1 }{ 2 }\\)
\n\"NCERT<\/p>\n

\"NCERT<\/p>\n

Question 10.
\n\\(\\frac { 2x-3 }{ ({ x }^{ 2 }-1)(2x+3) } \\)
\nSolution:
\nLet \\(\\frac{2 x-3}{(x-1)(x+1)(2 x+3)}=\\frac{\\mathrm{A}}{x-1}+\\frac{\\mathrm{B}}{x+1}+\\frac{\\mathrm{C}}{2 x+3}\\)
\n2x – 3 = A(x + 1)(2x + 3) + B(x – 1)(2x + 3) + C(x – 1)(x + 1) … (1)
\nPut x = 1 in (1), we get – 1 = 10A \u2234 A = \\(\\frac { – 1 }{ 10 }\\)
\nPut x = – 1 in (1), we get – 5 = – 2B \u2234 B = \\(\\frac { 5 }{ 2 }\\)
\nPut x = \\(\\frac { – 3 }{ 2 }\\) in (1), we get – 6 = \\(\\frac { 5 }{ 4 }\\)C \u2234 C = \\(\\frac { -24 }{ 5 }\\)
\n\"NCERT<\/p>\n

Question 11.
\n\\(\\frac { 5x }{ (x-1)({ x }^{ 2 }-4) } \\)
\nSolution:
\n\\(\\frac { 5x }{ (x-1)({ x }^{ 2 }-4) } =\\frac { 5x }{ (x+1)(x+2)(x-2) } \\)
\nLet
\n\\(\\frac { 5x }{ (x+1)(x+2)(x-2) }\\) = \\(\\frac{\\mathrm{A}}{x+1}+\\frac{\\mathrm{B}}{x+2}+\\frac{\\mathrm{C}}{x-2}\\)
\n5x = A(x + 2) (x – 2) + B(x + 1) (x – 2) + C (x + 1) (x + 2) … (1)
\nPut x = – 1 in (1), we get – 5 = – 3 A \u2234 A = \\(\\frac { 5 }{ 3 }\\)
\nPut x = – 2 in (1), we get – 10 = 4B \u2234 B = \\(\\frac { – 5 }{ 2 }\\)
\nPut x = 2 in (1), we get 10 = 12C \u2234 C = \\(\\frac { 5 }{ 6 }\\)
\n\"NCERT<\/p>\n

Question 12.
\n\\(\\frac { { x }^{ 3 }+x+1 }{ { x }^{ 2 }-1 } \\)
\nSolution:
\nDegree of Nr > Degree of Dr
\n\"NCERT<\/p>\n

\"NCERT<\/p>\n

Question 13.
\n\\(\\frac { 2 }{ (1-x)(1+{ x }^{ 2 }) } \\)
\nSolution:
\n\"NCERT<\/p>\n

Question 14.
\n\\(\\frac { 3x-1 }{ { (x+2) }^{ 2 } } \\)
\nSolution:
\n\\(\\frac { 3x-1 }{ { (x+2) }^{ 2 } } \\equiv \\frac { A }{ x+1 } +\\frac { B }{ { (x+2) }^{ 2 } } \\)
\n\u2234 3x – 1 = A(x + 2) + B … (i)
\nPut x = – 2 in (1), we get – 7 = B \u2234B = – 7
\nEquating the coefficients of x, we get A = 3
\n\u2234 \\(\\frac{3 x-1}{(x+2)^{2}}=\\frac{3}{(x+2)}-\\frac{7}{(x+2)^{2}}\\)
\n\\(\\int \\frac{3 x-1}{(x+2)^{2}} d x=3 \\int \\frac{1}{(x+2)} d x-7 \\int \\frac{1}{(x+2)^{2}} d x\\)
\n= 3 log|x+2| – 7(\\(\\left(\\frac{-1}{x+2}\\right)\\)) + C
\n= \\(3log|x+2|+\\frac { 7 }{ x+2 } +C\\)<\/p>\n

Question 15.
\n\\(\\frac { 1 }{ { x }^{ 4 }-1 } \\)
\nSolution:
\n\\(\\frac { 1 }{ { x }^{ 4 }-1 } \\)
\n= \\(\\frac{1}{\\left(x^{2}-1\\right)\\left(x^{2}+1\\right)}\\)
\n= \\(\\overline{(x-1)(x+1)\\left(x^{2}+1\\right)}\\)
\nLet
\n\\(\\frac { 1 }{ { x }^{ 4 }-1 } =\\frac { A }{ x+1 } +\\frac { B }{ x-1 } +\\frac { Cx+D }{ { x }^{ 2 }+1 } \\)
\n1 = A(x + 1)(x\u00b2 + 1) + B(x – 1)(x\u00b2 + 1) + (Cx + D)(x\u00b2 – 1)
\n1 = A(x + 1)(x\u00b2 + 1) + B(x – 1)(x\u00b2 + 1) + C(x\u00b3 – x) + D(x\u00b2 – 1) … (1)
\nPut x = 1 in (1), we get 1 = 4A
\n\u2234 A = \\(\\frac { 1 }{ 4 }\\)
\nPut x = -1 in (1), we get 1 = – 4B
\n\u2234 B = \\(\\frac { – 1 }{ 4 }\\)
\nEquating coefficients of x\u00b3 and constant terms, we get,
\nA + B + C = 0
\nA – B – D = 1
\n\u2234 C = 0 and D = \\(\\frac { – 1 }{ 2 }\\)
\n\\(\\frac{1}{x^{4}-1}=\\frac{1}{4(x-1)}-\\frac{1}{4(x+1)}-\\frac{1}{2\\left(x^{2}+1\\right)}\\)
\nIntegrating w.r.t. x, we get
\n\"NCERT<\/p>\n

\"NCERT<\/p>\n

Question 16.
\n\\(\\frac { 1 }{ x({ x }^{ n }+1) } \\)
\nSolution
\nMultiply numerator and denominator by xn-1<\/sup>
\n\"NCERT<\/p>\n

Question 17.
\n\\(\\frac { cosx }{ (1-sinx)(2-sinx) } \\)
\nSolution:
\nPut t = sinx, \\(\\frac { dt }{ dx }\\) = cosx, dt = cosx dx
\n\"NCERT<\/p>\n

Question 18.
\n\\(\\frac { \\left( { x }^{ 2 }+1 \\right) \\left( { x }^{ 2 }+2 \\right) }{ \\left( { x }^{ 2 }+3 \\right) \\left( { x }^{ 2 }+4 \\right) } \\)
\nSolution:
\n\"NCERT<\/p>\n

Question 19.
\n\\(\\frac { 2x }{ ({ x }^{ 2 }+1)({ x }^{ 2 }+3) } \\)
\nSolution:
\nput x\u00b2 = y
\nso that 2x dx = dy
\n\\(\\int \\frac{2 x}{\\left(x^{2}+1\\right)\\left(x^{2}+3\\right)} d x=\\int \\frac{d t}{(t+1)(t+3)}\\)
\nLet \\(\\frac{1}{(t+1)(t+3)}=\\frac{A}{t+1}+\\frac{B}{t+3}\\)
\n\u2234 1 = A(t + 3) + B(t + 1) … (1)
\nPut t = – 1 in (1), we get 1 = 2A \u2234 A = \\(\\frac { 1 }{ 2 }\\)
\nPut x = – 3 in (1), we get 1 = – 2B \u2234 B = \\(\\frac { – 1 }{ 2 }\\)
\n\"NCERT<\/p>\n

\"NCERT<\/p>\n

Question 20.
\n\\(\\frac { 1 }{ x({ x }^{ 4 }-1) } \\)
\nSolution:
\n\"NCERT<\/p>\n

Question 21.
\n\\(\\frac { 1 }{ { e }^{ x }-1 } \\)
\nSolution:
\n\"NCERT<\/p>\n

Question 22.
\nChoose the correct answer in each of the following :
\n\\(\\int { \\frac { xdx }{ (x-1)(x-2) } equals } \\)
\n(a) \\(log\\left| \\frac { { (x-1) }^{ 2 } }{ x-2 } \\right| + C\\)
\n(b) \\(log\\left| \\frac { { (x-2) }^{ 2 } }{ x-1 } \\right| + C\\)
\n(c) \\(log\\left| \\left( \\frac { x-{ 1 }^{ 2 } }{ x-2 } \\right) \\right| +C\\)
\n(d) log|(x – 1)(x – 2)| + C
\nSolution:
\nLet \\(\\frac{x}{(x-1)(x-2)}=\\frac{A}{(x-1)}+\\frac{B}{(x-2)}\\)
\n\u2234 x = A(x – 2) + B(x – 1) … (1)
\nPut x = 1 in (1), we get 1 = – A \u2234 A = – 1
\nPut x = 2 in (1), we get B = 1
\n\"NCERT<\/p>\n

Question 23.
\n\\(\\int { \\frac { dx }{ x({ x }^{ 2 }+1) } } \\) equals
\n(a) \\(log|x|-\\frac { 1 }{ 2 } log({ x }^{ 2 }+1)+C \\)
\n(b) \\(log|x|+\\frac { 1 }{ 2 } log({ x }^{ 2 }+1)+C \\)
\n(c) \\(-log|x|+\\frac { 1 }{ 2 } log({ x }^{ 2 }+1)+C\\)
\n(d) \\(\\frac { 1 }{ 2 } log|x|+log({ x }^{ 2 }+1)+C \\)
\nSolution:
\n\\(\\int { \\frac { dx }{ x({ x }^{ 2 }+1) } } \\) dx
\nLet
\n\\(\\frac { 1 }{ x\\left( { x }^{ 2 }+1 \\right) } =\\frac { A }{ x } +\\frac { Bx+C }{ { x }^{ 2 }+1 } \\)
\n\u21d2 1 = A(x\u00b2 + 1) + (Bx + C)x
\nEquating the coefficients of x\u00b2, x and con-stant term, we get A + B = 0, C = 0, A = 1
\nSolving the equations, we get A = 1, B = – 1, C = 0
\n\\(\\frac{1}{x\\left(x^{2}+1\\right)}=\\frac{1}{x}+\\frac{-x}{x^{2}+1}\\)
\n\\(\\int \\frac{1}{x\\left(x^{2}+1\\right)} d x=\\int \\frac{1}{x} d x-\\frac{1}{2} \\int \\frac{2 x}{x^{2}+1} d x\\)
\n= \\(\\log |x|-\\frac{1}{2} \\log \\left|x^{2}+1\\right|\\) + C<\/p>\n","protected":false},"excerpt":{"rendered":"

These NCERT Solutions for Class 12 Maths Chapter 7 Integrals Ex 7.5 Questions and Answers are prepared by our highly skilled subject experts. https:\/\/mcq-questions.com\/ncert-solutions-for-class-12-maths-chapter-7-ex-7-5\/ NCERT Solutions for Class 12 Maths Chapter 7 Integrals Exercise 7.5 Exercise 7.5 Class 12 Maths Solutions Question 1. Solution: Solution: let \u2261 \u2234 x = A(x + 2) + B(x …<\/p>\n

NCERT Solutions for Class 12 Maths Chapter 7 Integrals Ex 7.5<\/span> Read More »<\/a><\/p>\n","protected":false},"author":9,"featured_media":0,"comment_status":"closed","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"site-sidebar-layout":"default","site-content-layout":"default","ast-main-header-display":"","ast-hfb-above-header-display":"","ast-hfb-below-header-display":"","ast-hfb-mobile-header-display":"","site-post-title":"","ast-breadcrumbs-content":"","ast-featured-img":"","footer-sml-layout":"","theme-transparent-header-meta":"default","adv-header-id-meta":"","stick-header-meta":"default","header-above-stick-meta":"","header-main-stick-meta":"","header-below-stick-meta":"","spay_email":""},"categories":[3],"tags":[],"yoast_head":"\nNCERT Solutions for Class 12 Maths Chapter 7 Integrals Ex 7.5 - MCQ Questions<\/title>\n<meta name=\"robots\" content=\"index, follow, max-snippet:-1, max-image-preview:large, max-video-preview:-1\" \/>\n<link rel=\"canonical\" href=\"https:\/\/mcq-questions.com\/ncert-solutions-for-class-12-maths-chapter-7-ex-7-5\/\" \/>\n<meta property=\"og:locale\" content=\"en_US\" \/>\n<meta property=\"og:type\" content=\"article\" \/>\n<meta property=\"og:title\" content=\"NCERT Solutions for Class 12 Maths Chapter 7 Integrals Ex 7.5 - MCQ Questions\" \/>\n<meta property=\"og:description\" content=\"These NCERT Solutions for Class 12 Maths Chapter 7 Integrals Ex 7.5 Questions and Answers are prepared by our highly skilled subject experts. https:\/\/mcq-questions.com\/ncert-solutions-for-class-12-maths-chapter-7-ex-7-5\/ NCERT Solutions for Class 12 Maths Chapter 7 Integrals Exercise 7.5 Exercise 7.5 Class 12 Maths Solutions Question 1. 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