NCERT Solutions for Class 12 Maths Chapter 7 समाकलन Ex 7.5

These NCERT Solutions for Class 12 Maths Chapter 7 समाकलन Ex 7.5 Questions and Answers are prepared by our highly skilled subject experts.

NCERT Solutions for Class 12 Maths Chapter 7 समाकलन Ex 7.5

प्रश्न 1 से 21 तक के प्रश्नों में परिमेय फलनों का समाकलन कीजिए।
प्रश्न 1.
\(\frac{x}{(x+1)(x+2)}\)
हल:
आंशिक भिन्नों का प्रयोग करने पर,
माना
\(\frac{x}{(x+1)(x+2)}\) = \(\frac{A}{x+1}\) + \(\frac{B}{x+1}\)
या x = A(x + 2) + B(x + 1)
या
x = (A + B) x + 2A + B
दोनों पक्षों में x के गुणांकों तथा अचर पदों की तुलना करने पर,
A + B = 1, 2A + B = 0
इन समीकरणों को हल करने पर,
A = -1 तथा B = 2
∴ \(\frac{x}{(x+1)(x+2)}\) = \(\frac{-1}{x+1}\) + \(\frac{2}{x+2}\)
∴ \(\frac{x}{(x+1)(x+2)}\) dx = – ∫\(\frac{d x}{x+1}\) + ∫\(\frac{2}{x+2}\) dx
= – log|x + 1| + 2 log|x + 2| + C
= – log|x + 1| + log|x + 2|² + C
= log \(\frac{(x+2)^2}{|x+1|}\) + C
(∵ (x + 2)² > 0)

प्रश्न 2.
\(\frac{1}{x^2-9}\)
हल:
माना
\(\frac{1}{x^2-9}\) = \(\frac{1}{(x-3)(x+3)}\)
= \(\frac{A}{x-3}\) + \(\frac{B}{x+3}\)
या
1 = A(x + 3) + B(x – 3)
या
1 = (A + B)x + 3A – 3B
दोनों पक्षों में x के गुणांकों तथा अचर पदों की तुलना करने पर,
A + B = 0 तथा 3A – 3B = 1
इन समीकरणों को हल करने पर,
A = \(\frac{1}{6}\) B = –\(\frac{1}{6}\)
∴ ∫\(\frac{1}{x^2-9}\) dx = \(\frac{1}{6}\)∫\(\frac{d x}{x-3}\) – \(\frac{1}{6}\)∫\(\frac{d x}{x+3}\)
= \(\frac{1}{6}\)log|x – 3| – \(\frac{1}{6}\)log|x + 3| + C
= \(\frac{1}{6}\)log\(\left|\frac{x-3}{x+3}\right|\) + C

प्रश्न 3.
\(\frac{3 x-1}{(x-1)(x-2)(x-3)}\)
हल:
माना
\(\frac{3 x-1}{(x-1)(x-2)(x-3)}\)
= \(\frac{A}{x-1}\) + \(\frac{B}{x-2}\) + \(\frac{C}{x-3}\)
या 3x – 1 = A(x – 2)(x – 3) + B(x – 1)(x -3) + C(x – 1)(x – 2)
या 3x – 1 = A(x² – 5x + 6) + B(x² – 4x + 3) + C(x² – 3x + 2)
या 3x – 1 = (A + B + C)² + (- 5A – 4B – 3C)x + 6A + 3B + 2C
दोनों पक्षों में x के गुणांकों तथा अचर पदों की तुलना करने पर,
A + B + C = 0
-5A – 4B – 3C = 3
6A + 3B + 2C = -1
इन समीकरणों को हल करने पर,
\(\frac{3 x-1}{(x-1)(x-2)(x-3)}\) = \(\frac{1}{x-1}\) – \(\frac{5}{x-2}+\frac{4}{x-3}\)
∴ ∫\(\frac{3 x-1}{(x-1)(x-2)(x-3)}\)dx = ∫\(\frac{1}{x-1}\)dx – 5∫\(\frac{d x}{x-2}\) + ∫\(\frac{d x}{x-3}\)
log|x – 1| – 5 log|x – 2| + 4log|x – 3| + C

प्रश्न 4.
\(\frac{x}{(x-1)(x-2)(x-3)}\)
हल:
माना \(\frac{x}{(x-1)(x-2)(x-3)}\)
= \(\frac{A}{x-1}\) + \(\frac{B}{x-2}\) + \(\frac{C}{x-3}\)
या x = A(x – 2)(x – 3) + B(x – 1)(x – 3) + C(x – 1)(x – 2) ……………(1)
समीकरणा (1) में x = 1 रखने पर,
1 = A(1 – 2)(1 – 3) + 0 + 0
या 1 = A(-1)(-2) = 2A
∴A = \(\frac{1}{2}\)
समीकरणा (1) में x = 2 रखने पर,
2 = B(2 – 1)(2 – 3)
या 2 = B(1)(-1)
या 2 = -B
∴ B = – 2
समीकरणा (1) में x = 3 रखने पर,
3 = C(3 – 1)(3 – 2) = C(2) × 1
या 3 = 2C
C = \(\frac{3}{2}\)
∴ \(\frac{x}{(x-1)(x-2)(x-3)}\) = \(\frac{1}{2(x-1)}\) – \(\frac{2}{x-2}\) + \(\frac{3}{2(x-3)}\)
∫\(\frac{x}{(x-1)(x-2)(x-3)}\) dx
= \(\frac{1}{2}\)∫\(\frac{1}{2(x-1)}\) – 2∫\(\frac{2}{x-2}\) + \(\frac{3}{2}\) \(\frac{3}{2(x-3)}\)
= \(\frac{1}{2}\)log|x – 1||-2log|x – 2| + \(\frac{3}{2}\)log|x – 3|+ C

प्रश्न 5.
\(\frac{2 x}{x^2+3 x+2}\)
हल:
माना \(\frac{2 x}{x^2+3 x+2}\) = \(\frac{2 x}{(x+1)(x+2)}\)
= \(\frac{A}{(x+1)}\) + \(\frac{B}{(x+2)}\)
या 2x = A(x + 2) + B(x + 1) ……………….(1)
समीकरण (1) में x = -1 रखने पर,
2 × (-1) = A(-1 + 2) = A
या -2 = A
∴ A = -2
समीकरण (1) में x = -2 रखने पर,
2 × (-2) = A(-2 + 1) = -B
या -4 = -B
∴ B = 4
∴ \(\frac{2 x}{x^2+3 x+2}\) = \(\frac{-2}{x+1}\) + \(\frac{4}{x+2}\)
∴∫\(\frac{2 x}{x^2+3 x+2}\) = – 2∫\(\frac{d x}{x+1}\) + 4∫\(\frac{d x}{x+2}\)
= – 2log|x + 1| |4 + log|x + 2| + C
= 4 log|x + 2| – 2 log|x + 1| + C

प्रश्न 6.
\(\frac{1-x^2}{x(1-2 x)}\)
हल:
समाकल्य \(\frac{1-x^2}{x(1-2 x)}\) में अंश तथा हर की घात समान है, अंश को हर से भाग देने पर,

प्रश्न 7.
\(\frac{x}{\left(x^2+1\right)(x-1)}\)
हल:
माना \(\frac{x}{\left(x^2+1\right)(x-1)}\) = \(\frac{A}{x-1}\) + \(\frac{B x+C}{x^2+1}\)
या x = A(x² + 1) + (Bx + C)(x – 1)
या x = A(x² + 1) + (Bx² + Cx – Bx – C)
या x = (A + B)x² + (C – B)x + A – C
दोनों पक्षों में x, x² के गुणांकों तथा अचर पदों की तुलना करने पर,
A + B = 0, C – B = 1, A – C = 0
इन समीकरणों को हल करने पर,
A = \(\frac{1}{2}\) B = –\(\frac{1}{2}\) C = \(\frac{1}{2}\)

प्रश्न 8.
\(\frac{x}{(x-1)^2(x+2)}\)
हल:
माना \(\frac{x}{(x-1)^2(x+2)}\) = \(\frac{A}{x-1}\) + \(\frac{B}{(x-1)^2}\) + \(\frac{C}{x+2}\)
या x = A(x – 1)(x + 2) + B(x + 2) + C(x – 1)² …………….(1)
समीकरण (1) में x = 1 रखने पर,
1 = B(1 + 2) = 3B
∴ B = \(\frac{1}{3}\)
समीकरण (1) में x = -2 रखने पर,
-2 = C(-2 – 1)²
या -2 = C(-3)² = 9C
∴ C = – \(\frac{2}{9}\)
पुन: (1) से, x = A(x² + x – 2) + B(x + 2) + C(x² – 2x + 1)
या x = (A + C)x² + (A + B – 2C)x – 2A + 2B + C
दोनों पक्षों में x तथा x² के गुणांकों तथा अचर पदों की तुलना करने पर,
A + C = 0
A + B – 2C = 1,
-2A + 2B + C = 0
∵ B = \(\frac{1}{3}\), C = –\(\frac{2}{9}\)
∵ A + B – 2C = 1
तब A = 1 – B + 2C
= 1 – \(\frac{1}{3}\) – 2 × \(\frac{2}{9}\)
= \(\frac{2}{3}\) – \(\frac{4}{9}\) = \(\frac{6-4}{9}\) = \(\frac{2}{9}\)
∴ A = \(\frac{2}{9}\)
∴ \(\frac{x}{(x-1)^2(x+2)}\) = \(\frac{2}{9(x-1)}\) + \(\frac{1}{3(x-1)^2}\) – \(\frac{2}{9(x+2)}\)

प्रश्न 9.
\(\frac{3 x+5}{x^3-x^2-x+1}\)
हल:
\(\frac{3 x+5}{x^3-x^2-x+1}\) = \(\frac{3 x+5}{x^2(x-1)-(x-1)}\)
= \(\frac{3 x+5}{\left(x^2-1\right)(x-1)}\)
= \(\frac{3 x+5}{(x-1)^2(x+1)}\)
माना = \(\frac{3 x+5}{(x-1)^2(x+1)}\) = \(\frac{A}{x-1}\) + \(\frac{B}{(x-1)^2}\) + \(\frac{C}{(x+1)}\)
या 3x + 5 = A(x – 1)(x + 1) + B(x + 1) + C(x – 1)²
या 3x + 5 = A(x² – 1) + B(x – 1) + C(x² – 2x + 1)
या 3x + 5 = (A + C)x² + (B – 2C)x – A + B + C
दोनों पक्षों में x तथा x² के गुणांकों तथा अचर पदों की तुलना करने पर,
A + C = 0
B – 2C = 3, -A + B + C = 5
इन समीकरणों को हल करने पर,
A = –\(\frac{1}{2}\), B = 4, C = \(\frac{1}{2}\)

प्रश्न 10.
\(\frac{2 x-3}{\left(x^2-1\right)(2 x+3)}\)
हल:
\(\frac{2 x-3}{\left(x^2-1\right)(2 x+3)}\) = \(\frac{2 x-3}{(x-1)(x+1)(2 x+3)}\)
= \(\frac{A}{x-1}\) + \(\frac{B}{x+1}\) + \(\frac{C}{2 x+3}\)
या 2x – 3 = A(x + 1)(2x + 3) + B(x – 1)(2x + 3) + C(x – 1)(x + 1)
x = 1 रखने पर,
2 × 1 – 3 = A(1 + 1)(2 × 1 + 3)
या 2 – 3 = A × 2 × 5 = 10A
या -1 = 10A
या A = –\(\frac{1}{10}\)
x = – 1रखने पर,
2 × (-1) – 3 = B(-1 -1)(2 × (-1) + 3)
या – 2 – 3 = B(-2)(-2 + 3) = -2B
या 2B = 5 ∴ B = \(\frac{5}{2}\)
x = –\(\frac{3}{2}\) रखने पर,

प्रश्न 11.
\(\frac{5 x}{(x+1)\left(x^2-4\right)}\)
हल:
\(\frac{5 x}{(x+1)\left(x^2-4\right)}\) = \(\frac{5 x}{(x+1)(x-2)(x+2)}\)
= \(\frac{A}{x+1}\) + \(\frac{B}{x+2}\) + \(\frac{C}{x-2}\)
∴ 5x = A(x – 2)(x + 2) + B(x + 1)(x – 2) + C(x + 1)(x + 2)
x = -1 रखने पर,
5 × (-1) = A(-1 -2)(-1 + 2)
या -5 = A(-3)(1) = -3A
∴ A = \(\frac{5}{3}\)
x = -2 रखने पर,
5 × 2 = B(-2 + 1)(-2 – 2)
या -10 = B(-1)(-4) = 4B
∴ B = –\(\frac{10}{4}\) = –\(\frac{5}{2}\)
x = 2 रखने पर,
5 × 2 = C(2 + 1)(2 + 2)
या 10 = C(3)(4)
∴ C = \(\frac{10}{12}\) = \(\frac{5}{6}\)
∴ \(\frac{5 x}{(x+1)\left(x^2-4\right)}\) = \(\frac{5}{3(x+1)}\) – \(\frac{5}{2(x+2)}\) + \(\frac{5}{6(x-2)}\)
∴ ∫\(\frac{5 x}{(x+1)\left(x^2-4\right)}\)dx = \(\frac{5}{3}\)∫\(\frac{d x}{x+1}\) – \(\frac{5}{2}\)∫\(\frac{d x}{x+2}\) + \(\frac{5}{6}\)∫\(\frac{d x}{x-2}\)
= \(\frac{5}{3}\)log|x + 1| – \(\frac{5}{2}\)log|x + 2| + \(\frac{5}{6}\)log|x – 2| + C

प्रश्न 12.
\(\frac{x^3+x+1}{x^2-1}\)
हल:
दिया गया फलन विषम परिमेय फलन है।
माना x³ + x + 1 x² – 1 से भाग देने पर,
\(\frac{x^3+x+1}{x^2-1}\) = x + \(\frac{2 x+1}{x^2-1}\)
अब \(\frac{2 x+1}{x^2-1}\) = \(\frac{2 x+1}{(x-1)(x+1)}\)
= \(\frac{A}{x-1}\) + \(\frac{B}{x+1}\)
या 2x + 1 = A(x + 1) + B(x -1)
x = 1 रखने पर,
2 × 1 + 1 = A(1 + 1) + 0
या 3 = 2A
∴ A = \(\frac{3}{2}\)
x = -1 रखने पर,
2(-1) + 1 = B(-1 – 1) = – 2B
या – 2 + 1 = -2B
या -1 = – 2B
∴ B = \(\frac{1}{2}\)
∴ \(\frac{2 x+1}{x^2-1}\) = \(\frac{3}{2(x-1)}\) + \(\frac{1}{2(x+1)}\)
∴ ∫\(\frac{x^3+x+1}{x^2-1}\) dx = ∫x dx + \(\frac{3}{2}\)∫\(\frac{d x}{x-1}\) + \(\frac{1}{2}\)∫\(\frac{d x}{x+1}\)
= \(\frac{x^2}{2}\) + \(\frac{3}{2}\) log|x -1| + \(\frac{1}{2}\)log|x -1| + C
= \(\frac{x^2}{2}\) + \(\frac{1}{2}\) log|x -1| + \(\frac{3}{2}\)log|x -1| + C

प्रश्न 13.
\(\frac{2}{(1-x)\left(1+x^2\right)}\)
हल:
\(\frac{2}{(1-x)\left(1+x^2\right)}\) = \(\frac{A}{(1-x)}\) + \(\frac{B x+C}{1+x^2}\)
या 2 = A(1 – x² ) + (Bx + C)(1 – x)
या 2 = A + Ax² + Bx – Bx² + C – Cx
या 2 = (A – B)x² + (B – C)x + A + C
दोनों पक्षों में x तथा x² के गुणांकों तथा अचर पदों की तुलना करने पर,
A – B = 0, B – C = 0 A + C = 2
इस समीकरणों को हल करने पर,
A = 1, B = 1, C = 1
∴ ∫\(\frac{2}{(1-x)\left(1+x^2\right)}\) dx = ∫\(\frac{1}{1-x}\) dx + \(\frac{1}{2}\)∫\(\frac{2 x}{1+x^2}\)dx + ∫\(\frac{1}{1+x^2}\)dx
1 – x = t
-dx = dt
dx = -dt
∫\(\frac{d x}{1-x}\) = ∫\(\frac{d t}{t}\)
= – logt
= – log(1 – x)
∫\(\frac{2}{(1-x)\left(1+x^2\right)}\) dx = – log(1 – x) + \(\frac{1}{2}\) (1 + x² ) + tan^-1 x + C

प्रश्न 14.
\(\frac{3 x-1}{(x+2)^2}\)
हल:
\(\frac{3 x-1}{(x+2)^2}\) = \(\frac{A}{(x+2)}\) + \(\frac{B}{(x+2)^2}\)
या 3x – 1 = A(x + 2) + B
या 3x – 1 = Ax + 2A + B
दोनों पक्षों में x के गुणांकों तथा अचर पदों की तुलना करने पर,
A = 3 2A + B = -1
या
2 × 3 + B = -1
∴ B = -1-6 = -7
∴ \(\frac{3 x-1}{(x+2)^2}\) = \(\frac{3}{(x+2)}\) – \(\frac{7}{(x+2)^2}\)
∴ ∫\(\frac{3 x-1}{(x+2)^2}\)dx = ∫\(\frac{3}{(x+2)}\)dx – 7∫\(\frac{d x}{(x+2)^2}\)
= 3 log|x + 2| – 7(-\(\frac{1}{(x+2)}\)) + C
= 3 log|x + 2| + \(\frac{7}{x+2}\) + C

प्रश्न 15.
\(\frac{1}{x^4-1}\)
हल:
\(\frac{1}{x^4-1}\) = \(\frac{1}{(x-1)(x+1)\left(x^2+1\right)}\)
\(\frac{1}{(x-1)(x+1)\left(x^2+1\right)}\) = \(\frac{A}{x-1}\) + \(\frac{B}{x+1}\) + \(\frac{C x+D}{x^2+1}\)
या 1 = A(x + 1)(x² + 1) + B(x² + 1)(x – 1) + (Cx – D)(x -1)(x + 1)
या 1 = A(x³ + x² + x + 1) + B(x³ – x² + x – 1) + (Cx + D)(x -1)(x² – 1)
या 1 = A(x³ + x² + x + 1) + B(x³ – x² + x – 1) + (Cx³ – Cx + Dx² – D)
या 1 = (A + B + C)x³ + (A – B + D)x² + (A + B – C)x + A – B – D
दोनों पक्षों में x³, x² के गुणांकों तथा अचर पदों की तुलना करने पर,
A + B + C = 0
A – B + D = 0
A + B – C = 0
A – B – D = 1
इन समीकरणों को हल करने पर,
A = \(\frac{1}{4}\) B = \(\frac{1}{4}\), C = 0 D = –\(\frac{1}{2}\)
∴ \(\frac{1}{x^4-1}\) = \(\frac{1}{4(x-1)}\) – \(\frac{1}{4(x+1)}\) + \(\left[-\frac{1}{2\left(x^2+1\right)}\right]\)
= \(\frac{1}{4(x-1)}\) – \(\frac{1}{4(x+1)}\) + \(\frac{1}{2\left(x^2+1\right)}\)
∫\(\frac{1}{x^4-1}\)dx = \(\frac{1}{4}\)∫\(\frac{d x}{x-1}\) – \(\frac{1}{4}\)∫\(\frac{d x}{x+1}\) –
\(\frac{1}{4}\)∫\(\frac{d x}{x^2+1}\)
= \(\frac{1}{4}\)log|x – 1|- \(\frac{1}{4}\)log|x + 1| – \(\frac{1}{2}\)tan^-1 x + C
= \(\frac{1}{4}\)log\(\left|\frac{x-1}{x+1}\right|\) – \(\frac{1}{2}\)tan^-1 x + C

प्रश्न 16.
\(\frac{1}{x\left(x^n+1\right)}\)
हल:
\(\frac{1}{x\left(x^n+1\right)}\)
= \(\frac{x^{n-1}}{x^{n-1} \times x\left(x^n+1\right)}\)
(अंश तथा हर में x^{n – 1} का गुणा करने पर)
= \(\frac{x^{n-1}}{x^{n-1} \times x\left(x^n+1\right)}\)
∴ ∫\(\frac{1}{x\left(x^n+1\right)}\)dx = ∫\(\frac{x^{n-1}}{x^{n-1} \times x\left(x^n+1\right)}\)dx
माना xⁿ = t
तब nx^n-1dx = dt
या x^n-1dx = \(\frac{1}{n}\) dt
∴ ∫\(\frac{1}{x\left(x^n+1\right)}\) dx = \(\frac{1}{n}\) dt
अब ∫\(\int \frac{d t}{t(t+1)}\) = \(\frac{A}{t}\) + \(\frac{B}{t+1}\)
या 1 = A(t + 1) + Bt
या 1 = (A + B)t + A
दोनों पक्षों में t के गुणांकों तथा अचर पदों की तुलना करने पर,
A + B = 0 तथा A = 1
∴ B = -1
तब \(\frac{1}{t(t+1)}\) = \(\frac{1}{t}\) – \(\frac{1}{t+1}\)
तब ∫\(\frac{d t}{t(t+1)}\) = ∫\(\frac{1}{t}\) dt – ∫\(\frac{d t}{t+1}\)
= log|t| – log|t + 1| + C
∴ \(\frac{1}{n}\)∫\(\frac{d t}{t(t+1)}\) = \(\frac{1}{n}\)log|t| – \(\frac{1}{n}\)log|t + 1| + C
या ∫\(\frac{1}{x\left(x^n+1\right)}\)dx = \(\frac{1}{n}\)log|xⁿ| – \(\frac{1}{n}\)log|xⁿ + 1| + C
= \(\frac{1}{n}\)log\(\left|\frac{x^n}{x^n+1}\right|\) + C

प्रश्न 17.
\(\frac{\cos x}{(1-\sin x)(2-\sin x)}\)
हल:
∫\(\frac{\cos x}{(1-\sin x)(2-\sin x)}\) dx
माना sin x = t, तब cosx dx = dt
∴ ∫\(\frac{\cos x}{(1-\sin x)(2-\sin x)}\) = ∫\(\frac{d t}{(1-t)(2-t)}\)
अब \(\frac{1}{(1-t)(2-t)}\) = \(\frac{A}{1-t}\) + \(\frac{B}{2-t}\)
या 1 = A(2 – t) + B(1 – t)
या 1 = 2A + B – (A + B)t
दोनों पक्षों में t के गुणांकों तथा अचर पदों की तुलना करने पर,
A + B = 0 2A + B = 1
इन समीकरणों को हल करने पर, A = 1 B = -1
∴ \(\frac{1}{(1-t)(2-t)}\) = \(\frac{1}{1-t}\) – \(\frac{1}{2-t}\)
∴ ∫\(\frac{d t}{(1-t)(2-t)}\) = ∫\(\frac{1}{1-t}\) dt – ∫\(\frac{1}{2-t}\) dt
= – log|1 – t| – (-log|2 – t|) + C
= – log|1 – t| + log|2 – t| + C
∴ ∫\(\frac{\cos x}{(1-\sin x)(2-\sin x)}\) dx = – log|1 – sinx| + log|2 – sinx| + C
= log\(\left|\frac{2-\sin x}{1-\sin x}\right|\) + C

प्रश्न 18.
\(\frac{\left(x^2+1\right)\left(x^2+2\right)}{\left(x^2+3\right)\left(x^2+4\right)}\)
हल:
दिया गया फलन विषम परिमेय फलन है। अब x² रखने

प्रश्न 19.
\(\frac{2 x}{\left(x^2+1\right)\left(x^2+3\right)}\)
हल:
माना I = ∫\(\frac{2 x}{\left(x^2+1\right)\left(x^2+3\right)}\)dx
x² = t 2x dx = dt
∴ I = ∫\(\frac{2 x}{\left(x^2+1\right)\left(x^2+3\right)}\)dx
= ∫\(\frac{d t}{(t+1)(t+3)}\) …………(1)
अब \(\frac{1}{(t+1)(t+3)}\) = \(\frac{A}{t+1}\) + \(\frac{B}{t+3}\)
या 1 = A(t + 3) + B(t + 1)
1 = (A + B)t + 3A + B
दोनों पक्षों में t के गुणांकों तथा अचर पदों की तुलना करने पर,
A + B = 0 3A + B = 1
इन समीकरणों को हल करने पर,
A = \(\frac{1}{2}\) तथा B = –\(\frac{1}{2}\)
∴ ∫\(\frac{1}{(t+1)(t+3)}\) dt = \(\frac{1}{2}\)∫\(\frac{d t}{(t+1)}\) – \(\frac{1}{2}\)∫\(\frac{d t}{t+3}\)
= \(\frac{1}{2}\)log|x + 1| – \(\frac{1}{2}\)log|t + 3| + C
= \(\frac{1}{2}\)log\(\left|\frac{t+1}{t+3}\right|\) + C
= \(\frac{1}{2}\)log\(\left|\frac{x^2+1}{x^2+3}\right|\) + C

प्रश्न 20.
\(\frac{1}{x\left(x^4-1\right)}\)
हल:
माना I = ∫\(\frac{1}{x\left(x^4-1\right)}\) dx
= \(\frac{1}{4}\)∫\(\frac{1}{x\left(x^4-1\right)}\) dx
x⁴ = t तब 4x³ dx = dt
∴ ∫\(\frac{d t}{t(t-1)}\)
अब \(\frac{1}{t(t-1)}\) = \(\frac{A}{t}\) + \(\frac{B}{t-1}\)
1 = A(t – 1) + Bt
1 = (A + B)t – A
दोनों पक्षों में t के गुणांकों तथा अचर पदों की तुलना करने पर
A + B = 0, A = -1
∴ B = -1
∴ I = \(\frac{1}{4}\)∫\(\frac{d t}{t(t-1)}\)
= – \(\frac{1}{4}\)∫\(\frac{1}{t}\) + \(\frac{1}{4}\)∫\(\frac{d t}{t-1}\)
= – \(\frac{1}{4}\) log|t| + \(\frac{1}{4}\)log|t – 1| + C
= \(\frac{1}{4}\)log\(\left|\frac{t-1}{t}\right|\) + C
= \(\frac{1}{4}\)log\(\left|\frac{x^4-1}{x^4}\right|\) + C

प्रश्न 21.
\(\frac{1}{\left(e^x-1\right)}\)
हल:
माना I = ∫\(\frac{dx}{\left(e^x-1\right)}\) = ∫\(\frac{e^x d x}{e^x\left(e^x-1\right)}\)
e^x = t e^xdx = dt
∴ I = ∫\(\frac{d t}{t(t-1)}\)
अब \(\frac{1}{t(t-1)}\) = \(\frac{A}{t}\) + \(\frac{B}{t-1}\)
⇒ 1 = A(t – 1) + Bt
t = 0 रखने पर, A = -1
t = 1 रखने पर, B = -1
∴ ∫\(\frac{1}{t(t-1)}\) = \(-\frac{1}{t}\) + \(\frac{1}{t-1}\)
∴ ∫\(\frac{1}{t(t-1)}\) dt = – ∫\(-\frac{1}{t}\)dt + ∫\(\frac{d t}{t-1}\)
= -log|t| + log|t – 1| + C
= log\(\left|\frac{t-1}{t}\right|\) + C
= log\(\left|\frac{e^x-1}{e^x}\right|\) + C

प्रश्न 22.
∫\(\frac{x d x}{(x-1)(x-2)}\) बराबर है:
(A) log\(\left|\frac{(x-1)^2}{x-2}\right|\) + C
(B) log\(\left|\frac{(x-2)^2}{x-1}\right|\) + C
(C) log\(\left|\left(\frac{x-1}{x-2}\right)^2\right|\) + C
(D) log|(x – 1)(x – 2)| + C
हल:
माना
\(\frac{x}{(x-1)(x-2)}\) = \(\frac{A}{(x-1)}\) + \(\frac{B}{(x-2)}\)
x = A(x – 2) + B (x – 1)
x = 1 रखने पर
1 = A(1 – 2) = -A
A = -1
x = 2 रखने पर
2 = B(2 – 1) = B
∴ B = 2
∴ \(\frac{x}{(x-1)(x-2)}\) = \(\frac{-1}{x-1}\) + \(\frac{2}{x-2}\)
∴ ∫\(\frac{x}{(x-1)(x-2)}\) dx
= – ∫\(\frac{d x}{x-1}\) + 2∫\(\frac{d x}{x-2}\)
= – log|x – 1| + 2 log|x – 2| + C
= – log|x – 1| + log(x – 2)² + C
= log\(\left|\frac{(x-2)^2}{x-1}\right|\) + C
अतः विकल्प (B) सही है।

प्रश्न 23.
∫\(\frac{d x}{x\left(x^2+1\right)}\)बराबर है:
(A) log|x| – \(\frac{1}{2}|\)log(x² + 1) + C
(B) log|x| + \(\frac{1}{2}\)log(x² + 1) + C
(C) -log|x| + \(\frac{1}{2}\)log(x² + 1) + C
(D) \(\frac{1}{2}\)log|x|log(x² + 1) + C
हल: