NCERT Solutions for Class 12 Maths Chapter 7 समाकलन Ex 7.4

These NCERT Solutions for Class 12 Maths Chapter 7 समाकलन Ex 7.4 Questions and Answers are prepared by our highly skilled subject experts.

NCERT Solutions for Class 12 Maths Chapter 7 समाकलन Ex 7.4

प्रश्न 1 से 23 तक के फलनों का समाकलन कीजिए:
प्रश्न 1.
\(\frac{3 x^2}{x^6+1}\)
हल:
I = ∫\(\frac{3 x^2}{x^6+1}\) dx
= ∫\(\frac{3 x^2}{\left(x^3\right)^2+1}\) dx
माना x³ = t
तब
3x² dx = dt
∴ I = ∫\(\frac{d t}{t^2+1}\)
= tan^-1 t + C
= tan^-1 x³ + C

प्रश्न 2.
\(\frac{1}{\sqrt{1+4 x^2}}\)
हल:

प्रश्न 3.
\(\frac{1}{\sqrt{(2-x)^2+1}}\)
हल:
I = ∫\(\frac{1}{\sqrt{(2-x)^2+1}}\) dx
माना 2 – x = t
तब
– dx = dt या dx = – dt
∴ I = ∫\(\frac{-d t}{\sqrt{t^2+1}}\)
= – ∫\(\frac{-d t}{\sqrt{t^2+1}}\)
= – log | t + \(\sqrt{t^2+1}\) + C
= – log |2 – x + \(\sqrt{(2-x)^2+1}\)| + C
= – log|2 – x + \(\sqrt{x^2-4 x+4+1}\)| + C
= – log|2 – x + \(\sqrt{x^2-4 x+5}\)| + C
= log|\(\frac{1}{2-x+\sqrt{x^2-4 x+5}}\)| + C

प्रश्न 4.
\(\frac{1}{\sqrt{9-25 x^2}}\)
हल:
I = ∫\(\frac{1}{\sqrt{9-25 x^2}}\) dx
= \(\frac{1}{5}\) ∫\(\frac{d x}{\sqrt{\frac{9}{25}-x^2}}\)
= \(\frac{1}{5}\) ∫\(\frac{d x}{\sqrt{\left(\frac{3}{5}\right)^2-x^2}}\)
= \(\frac{1}{5}\) sin^-1 \(\frac{x}{3 / 5}\) + C
= \(\frac{1}{5}\) sin^-1 \(\frac{5 x}{3}\) + C

प्रश्न 5.
\(\frac{3 x}{1+2 x^4}\)
हल:
I = \(\frac{3}{2}\)∫\(\frac{x}{\frac{1}{2}+x^4}\) dx
= \(\frac{3}{2}\)∫\(\frac{x}{\frac{1}{2}+\left(x^2\right)^2}\) dx
माना
x² = t
तब
2x dx = dt
या
x dx = dt/2

प्रश्न 6.
\(\frac{x^2}{1-x^6}\)
हल:
I = ∫\(\frac{x^2}{1-\left(x^3\right)^2}\) dx
माना
x³ = t
तब
3x² dx = dt x² dx = \(\frac{1}{3}\) dt
∴ I = \(\frac{1}{3}\)∫\(\int \frac{d t}{1-t^2}\)
= \(\frac{1}{3}\) × \(\frac{1}{2 \times 1}\) log \(\left|\frac{1+t}{1-t}\right|\) + C
= \(\frac{1}{6}\) log \(\left|\frac{1+x^3}{1-x^3}\right|\) + C

प्रश्न 7.
\(\frac{x-1}{\sqrt{x^2-1}}\)
हल:
I = ∫\(\frac{x-1}{\sqrt{x^2-1}}\) dx
= ∫\(\frac{x}{\sqrt{x^2-1}}\) dx – ∫\(\int \frac{d x}{\sqrt{x^2-1}}\)
I = I₁ – I₂
I₁ = ∫\(\frac{x}{\sqrt{x^2-1}}\) dx
माना
x² – 1 = t
तब
2x dx = dt
या
x dx = \(\frac{1}{2}\) dt
∴ I₁ = \(\frac{1}{2}\)∫\(\frac{d t}{t^{1 / 2}}\) = \(\frac{1}{2}\)∫t^-1/2 dt
= \(\frac{1}{2}\)\(\frac{t^{-1 / 2+1}}{(-1 / 2+1)}\) + C₁
= \(\frac{1}{2}\)\(\frac{t^{1 / 2}}{1 / 2}\) + C₁
= \(\frac{1}{2}\) × 2t^1/2 + C₁
= t^1/2 + C₁
= \(\sqrt{x^2-1}\) + C₁
I² = ∫\(\frac{d x}{\sqrt{x^2-1}}\)
= log | x + \(\sqrt{x^2-1}\)|+ C₂
∴ I₁ तथा I₂ के मान (1) में रखने पर,
I = \(\sqrt{x^2-1}\) + C₁ – log | x + \(\sqrt{x^2-1}\)| + C2
= \(\sqrt{x^2-1}\) – log | x + \(\sqrt{x^2-1}\)| + C₁ + C₂
= \(\sqrt{x^2-1}\) – log | x + \(\sqrt{x^2-1}\)| + C
(∵ C = C₁ + C₂)

प्रश्न 8.
\(\frac{x^2}{\sqrt{x^6+a^6}}\)
हल:
I = ∫\(\frac{x^2}{\sqrt{\left(x^3\right)^2+\left(a^3\right)^2}}\) dx
माना
x³ = t
तब 3x² dx = dt
या x² dx = \(\frac{1}{3}\)dt
∴ I = \(\frac{1}{3}\)∫\(\frac{d t}{\sqrt{t^2+\left(a^3\right)^2}}\)
= \(\frac{1}{3}\)log|t + \(\sqrt{t^2+a^6}\)| + C
∴ I = \(\frac{1}{3}\)log|x³ + \(\sqrt{x^6+a^6}\)| + C

प्रश्न 9.
\(\frac{\sec ^2 x}{\sqrt{\tan ^2 x+4}}\)
हल:
I = ∫\(\frac{\sec ^2 x}{\sqrt{\tan ^2 x+4}}\) dx
माना
tan x = t
तब
sec² x dx = dt
∴ I = ∫\(\frac{d t}{\sqrt{t^2+4}}\)
= log|t + \(\sqrt{t^2+4}\)| + C
log | tan x + \(\sqrt{\tan ^2 x+4}\)| + C

प्रश्न 10.
\(\frac{1}{\sqrt{x^2+2 x+2}}\)
हल:

प्रश्न 11.
\(\frac{1}{9 x^2+6 x+5}\)
हल:

प्रश्न 12.
\(\frac{1}{\sqrt{7-6 x-x^2}}\)
हल:

प्रश्न 13.
\(\frac{1}{\sqrt{(x-1)(x-2)}}\)
हल:

प्रश्न 14.
\(\frac{1}{\sqrt{8+3 x-x^2}}\)
हल:

प्रश्न 15.
\(\frac{1}{\sqrt{(x-a)(x-b)}}\)
हल:

प्रश्न 16.
\(\frac{4 x+1}{\sqrt{2 x^2+x-3}}\)
हल:
I = ∫\(\frac{4 x+1}{\sqrt{2 x^2+x-3}}\)
माना 2x² + x – 3 = t,
तब (4x + 1) dx = dt
∴ I = ∫\(\frac{4 x+1}{\sqrt{2 x^2+x-3}}\) dx
= ∫\(\frac{d t}{t^{1 / 2}}\)
= ∫t^-1/2 dt
= \(\frac{t^{-1 / 2+1}}{-1 / 2+1}\) + C
= \(\frac{t^{1 / 2}}{1 / 2}\) + C
= 2t^-1/2 + C
= 2\(\sqrt{2 x^2+x-3}\) + C

प्रश्न 17.
\(\frac{x+2}{\sqrt{x^2-1}}\)
हल:
I = ∫\(\frac{x+2}{\sqrt{x^2-1}}\) dx
= ∫\(\frac{x}{\sqrt{x^2-1}}\) dx + ∫\(\frac{2}{\sqrt{x^2-1}}\) dx
I = I₁ + I₂ ………….(1)
I₁ = ∫\(\frac{x}{\sqrt{x^2-1}}\) dx
माना x² – 1 = t तब 2x dx = dt
∴ I₁ = \(\frac{1}{2}\)∫\(\frac{d t}{t^{1 / 2}}\)
= \(\frac{1}{2}\)∫t^-1/2 dt
= \(\frac{1}{2}\)\(\frac{t^{-1 / 2+1}}{-\frac{1}{2}+1}\) + C₁
I₁ तथा I₂ के मान (1) में रखने पर,
I₁ = \(\sqrt{x^2-1}\) + C₁ + 2log|x + \(\sqrt{x^2-1}\)| + C₂
I₁ = \(\sqrt{x^2-1}\) + 2log|x + \(\sqrt{x^2-1}\)| + C₁ + C₂
= \(\sqrt{x^2-1}\) + 2log|x + \(\sqrt{x^2-1}\)| + C

प्रश्न 18.
\(\frac{5 x-2}{1+2 x+3 x^2}\)
हल:
I = ∫\(\frac{5 x-2}{1+2 x+3 x^2}\) dx
माना संख्याएँ A तथा B इस प्रकार हैं कि
5x – 2 = A \(\frac{d}{d x}\) (1 + 2x + 3x² ) + B
या 5x – 2 = A(2 + 6x) + B
या 5x – 2 = 6Ax + 2A + B
दोनों पक्षों में x के गुणांकों तथा अचर पदों की तुलना करने पर,
6A = 5 तथा 2A + B = 2
∴ A = \(\frac{5}{6}\) तथा 2 × \(\frac{5}{6}\) + B = 2
या B = – 2 – \(\frac{5}{3}\) = \(\frac{11}{3}\)
∴ I = \(\frac{5}{6}\) ∫\(\frac{2+6 x}{1+2 x+3 x^2}\)dx – \(\frac{11}{3}\) ∫\(\frac{d x}{1+2 x+3 x^2}\)
I = \(\frac{5}{6}\)I₁ – \(\frac{11}{3}\)I₂ (माना) ……………..(1)
अब I₁ = ∫\(\frac{2+6 x}{1+2 x+3 x^2}\) dx
माना 1 + 2x + 3x² = t
तब (2 + 6x)dx = dt
∴ I₁ = ∫\(\frac{d t}{t}\)
= log|t| | C₁
= log|1 + 2x + 3x²| + C₁

प्रश्न 19.
\(\frac{6 x+7}{\sqrt{(x-5)(x-4)}}\)
हल:
I = ∫\(\frac{6 x+7}{\sqrt{(x-5)(x-4)}}\) dx
= ∫\(\frac{6 x+7}{\sqrt{x^2-9 x+20}}\) dx
माना संख्याएँ A तथा B इस प्रकार हैं कि
6x + 7 = A \(\frac{d}{d x}\) (x² – 9x + 20) + B
या 6x + 7 = A(2x – 9) + B
या 6x + 7 = 2Ax – 9A + B
दोनों पक्षों में x के गुणांकों तथा अचर पदों की तुलना करने पर,
2A = 6 तथा -9A + B = 7
A = 3 तथा – 27 + B = 7
या B = 27 + 7 = 34

प्रश्न 20.
\(\frac{x+2}{\sqrt{4 x-x^2}}\)
हल:

प्रश्न 21.
\(\frac{x+2}{\sqrt{x^2+2 x+3}}\)
हल:

प्रश्न 22.
\(\frac{x+3}{x^2-2 x-5}\)
हल:

प्रश्न 23.
∫\(\frac{5 x+3}{\sqrt{x^2+4 x+10}}\)
हल:
I = ∫\(\frac{5 x+3}{\sqrt{x^2+4 x+10}}\) dx
माना संख्याएँ A तथा B इस प्रकार हैं कि
5x + 3 = A \(\frac{d}{d x}\) (x² + 4x + 10) + B
या 5x + 3 = A(2x + 4) + B
या 5x + 3 = 2Ax + 4A + B
दोनों पक्षों में x के गुणांकों तथा अचर पदों की तुलना करने पर,
2A = 5 तथा 4A + B = 3
⇒ A = \(\frac{5}{2}\) तथा 4 × \(\frac{5}{2}\) + B = 3
या 10 + B = 3 या B = 3 – 10 = – 7
∴ I = \(\frac{5}{2}\)∫\(\frac{2 x+4}{\sqrt{x^2+4 x+10}}\) dx – 7∫\(\frac{d x}{\sqrt{x^2+4 x+10}}\)
या I = \(\frac{5}{2}\)I₁ – 7I₂(माना) …………….(1)
अब I₁ = ∫\(\frac{2 x+4}{\sqrt{x^2+4 x+10}}\) dx
माना x² + 4x + 10 = t
तब (2x + 4)dx = dt

प्रश्न 24 एवं 25 में सही उत्तर का चयन कीजिए।
प्रश्न 24.
∫\(\frac{d x}{x^2+2 x+2}\)बराबर है:
(A) xtan^-1(x + 1) + C
(B) tan^-1(x + 1) + C
(C) (x + 1)sin^-1 x + C
(D) tan^-1 x + C
हल:
I = ∫\(\frac{d x}{x^2+2 x+2}\)
= ∫\(\frac{d x}{x^2+2 x+1+1}\) = ∫\(\frac{d x}{(x+1)^2+1^2}\) = tan^-1\(\left(\frac{x+1}{1}\right)\) + C
= tan^-1(x + 1) + C
अतः विकल्प (B) सही है।

प्रश्न 25.
∫\(\frac{d x}{\sqrt{9 x-4 x^2}}\)बराबर है:
हल: