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 <\/p>\nQuestion 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 <\/p>\n
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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 <\/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 <\/p>\n
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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 <\/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 <\/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 <\/p>\n
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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 <\/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 <\/p>\n
Question 12. \n\\(\\frac { { x }^{ 3 }+x+1 }{ { x }^{ 2 }-1 } \\) \nSolution: \nDegree of Nr > Degree of Dr \n <\/p>\n
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Question 13. \n\\(\\frac { 2 }{ (1-x)(1+{ x }^{ 2 }) } \\) \nSolution: \n <\/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 <\/p>\n
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Question 16. \n\\(\\frac { 1 }{ x({ x }^{ n }+1) } \\) \nSolution \nMultiply numerator and denominator by xn-1<\/sup> \n <\/p>\nQuestion 17. \n\\(\\frac { cosx }{ (1-sinx)(2-sinx) } \\) \nSolution: \nPut t = sinx, \\(\\frac { dt }{ dx }\\) = cosx, dt = cosx dx \n <\/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 <\/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 <\/p>\n
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Question 20. \n\\(\\frac { 1 }{ x({ x }^{ 4 }-1) } \\) \nSolution: \n <\/p>\n
Question 21. \n\\(\\frac { 1 }{ { e }^{ x }-1 } \\) \nSolution: \n <\/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 <\/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 \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n\t \n\t \n\t \n