RD Sharma Class 12 Ex 19.5 Solutions Chapter 19 Indefinite Integrals

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Textbook	NCERT
Class	Class 12th
Subject	Maths
Chapter	19
Exercise	19.5
Category	RD Sharma Solutions

Table of Contents

RD Sharma Class 12 Ex 19.5 Solutions Chapter 19 Indefinite Integrals

Question 1. $\int\frac{x+1}{\sqrt{2x+3}}dx$

Solution:

Given integral, $\int\frac{x+1}{\sqrt{2x+3}}dx$
On Multiplying and dividing with 2, we get
⇒ $\frac{1}{2}\int\frac{2(x+1)}{\sqrt{2x+3}}dx$
⇒ $\frac{1}{2}\int\frac{(2x+3)-1}{\sqrt{2x+3}}dx$
⇒ $\frac{1}{2}(\int\sqrt{2x+3}dx-\int\frac{1}{\sqrt{2x+3}}dx)$
⇒ $\frac{1}{2}(\int(2x+3)^\frac{1}{2}dx-\int(2x+3)^\frac{-1}{2}dx)$
By using the formula,
$\int (ax+b)^ndx=\frac{(ax+b)^{n+1}}{a(n+1)}+c$ [where c is any arbitrary constant]
We get
⇒ $\frac{1}{2}[\frac{(2x+3)^{\frac{1}{2}+1}}{2(1/2 +1)}-\frac{(2x+3)^{\frac{-1}{2}+1}}{2(-1/2 +1)}]+c$
⇒ $\frac{1}{2}[\frac{(2x+3)^{\frac{3}{2}}}{3}-\frac{(2x+3)^{\frac{1}{2}}}{1}]+c$
⇒ $\frac{1}{6}(2x+3)^{\frac{3}{2}}-\frac{1}{2}(2x+3)^{\frac{1}{2}}+c$
⇒ $\frac{1}{2}(2x+3)^\frac{1}{2}[\frac{2x+3}{3}-1]+c$

Question 2. $\int x\sqrt{x+2}dx$

Solution:

Given integral, $\int x\sqrt{x+2}dx$
Let x + 2 =t ⇒ x = t – 2
On differentiating on both sides,
dx = dt
On substituting it in given integral, we get
⇒ $\int (t-2)\sqrt{t}dt$
⇒ $\int (t^\frac{3}{2}-2t^\frac{1}{2})dt$
We know that, $\int x^ndx=\frac{x^{n+1}}{n+1}+c$ [where c is any arbitrary constant]
⇒ $\frac{t^{\frac{3}{2}+1}}{\frac{3}{2}+1}-\frac{2t^{\frac{1}{2}+1}}{\frac{1}{2}+1}+c$
⇒ $\frac{2}{5}t^\frac{5}{2}-\frac{4}{3}t^\frac{3}{2}+c$
Replacing x in terms of t
⇒ $\frac{2}{5}(x+2)^\frac{5}{2}-\frac{4}{3}(x+2)^\frac{3}{2}+c$
⇒ $(x+2)^\frac{3}{2}[{\frac{2(x+2)}{5}}-\frac{4}{3}]+c$

Question 3. $\int\frac{x-1}{\sqrt{x+4}}dx$

Solution:

Given integral, $\int\frac{x-1}{\sqrt{x+4}}dx$
⇒ $\int\frac{x+4-5}{\sqrt{x+4}}dx$
⇒ $\int(\sqrt{x+4}-\frac{5}{\sqrt{x+4}})dx$
⇒ $\int(x+4)^\frac{1}{2}dx-5\int(x+4)^\frac{-1}{2}dx$
By using the formula,
$\int(ax+b)^ndx=\frac{(ax+b)^{n+1}}{a(n+1)}+ c$ [where c is any arbitrary constant]
We get
⇒ $\frac{(x+4)^{\frac{1}{2}+1}}{\frac{3}{2}}-5\frac{(x+4)^{\frac{-1}{2}+1}}{\frac{1}{2}}+c$
⇒ $2\frac{(x+4)^{\frac{3}{2}}}{3}-10(x+4)^\frac{1}{2}+c$
⇒ $(x+4)^\frac{1}{2}(\frac{2}{3}(x+4)-10)+c$

Question 4. $\int(x+2)\sqrt{3x+5}dx$

Solution:

Given integral, $\int(x+2)\sqrt{3x+5}dx$
Let 3x + 5 = t
⇒ x = (t – 5)/3
On differentiating both sides,
dx = dt/3
On replacing the x terms with t,
⇒ $\int(\frac{t-5}{3}+2)\sqrt{t}\frac{dt}{3}$
⇒ $\frac{1}{9}\int(t+1)\sqrt{t}dt$
⇒ $\frac{1}{9}\int t^{{3}/{2}}dt +\frac{1}{9}\int t^{{1}/{2}}dt$
By using the formula,
$\int x^ndx=\frac {x^{n+1}}{n+1}+c$ [where c is any arbitrary constant]
We get
⇒ $\frac{1}{9}(\frac{t^{\frac{3}{2}+1}}{\frac{3}{2}+1}+\frac{t^{\frac{1}{2}+1}}{\frac{1}{2}+1})+c$
⇒ $\frac{2}{9}(\frac{t^\frac{5}{2}}{5}+\frac{t^\frac{3}{2}}{3})+c$
On replacing t with x terms
⇒ $\frac{2}{9}(\frac{{(3x+5)}^\frac{5}{2}}{5}+\frac{{(3x+5)}^\frac{3}{2}}{3})+c$
⇒ $\frac{2}{9}{(3x+5)^{\frac{3}{2}}}(\frac{9x+20}{15})+c$
⇒ $\frac{2}{135}(9x+20)(3x+5)^\frac{3}{2}+c$

Question 5. $\int\frac{2x+1}{\sqrt{3x+2}}dx$

Solution:

Given integral, $\int\frac{2x+1}{\sqrt{3x+2}}dx$
On multiplying and dividing it with 3
⇒ $\frac{1}{3}\int\frac{6x+3}{\sqrt{3x+2}}dx$
⇒ $\frac{1}{3}\int\frac{(6x+4)-1}{\sqrt{3x+2}}dx$
⇒ $\frac{1}{3}\int(\frac{2(3x+2)}{\sqrt{3x+2}}-\frac{1}{\sqrt{3x+2}}) dx$
⇒ $\frac{1}{3}(\int{2\sqrt{3x+2}}dx-\int(3x+2)^\frac{-1}{2}dx)$
By using the formula,
$\int(ax+b)^ndx=\frac{(ax+b)^{n+1}}{a(n+1)}+c$ [where c is any arbitrary constant]
We get
⇒ $\frac{1}{3}[\frac{2(3x+2)^{\frac{3}{2}}}{\frac{9}{2}}-\frac{(3x+2)^{\frac{1}{2}}}{\frac{3}{2}}]+c$
⇒ $\frac{1}{3}[\frac{4}{9}{(3x+2)^\frac{3}{2}}-\frac{2}{3}(3x+2)^\frac{1}{2}]+c$
⇒ $\frac{4}{27}(3x+2)^\frac{3}{2}-\frac{2}{9}(3x+2)^{\frac{1}{2}}+c$
⇒ $\sqrt{3x+2}[\frac{4}{27}(3x+2)-\frac{2}{9}]+c$
⇒ $\frac{2}{27}(6x+1)\sqrt{3x+2}+c$

Question 6. $\int\frac{3x+5}{\sqrt{7x+9}}dx$

Solution:

Given integral, $\int\frac{3x+5}{\sqrt{7x+9}}dx$
Let 7x + 9 = t
⇒ x = (t – 9)/7
On differentiating both sides,
dx = dt/7
On replacing x terms with t
⇒ $\int(\frac{3(\frac{t-9}{7})+5}{\sqrt{t}})\frac{dt}{7}$
⇒ $\frac{1}{49}\int\frac{3t+8}{\sqrt{t}}dt$
⇒ $\frac{1}{49}\int(3\sqrt{t}+8t^\frac{-1}{2})dt$
By using the formula,
$\int x^{n}dx=\frac{x^{n+1}}{n+1}+c$ [where c is any arbitrary constant]
⇒ $\frac{1}{49}(\frac{3t^\frac{3}{2}}{\frac{3}{2}}+\frac{8t^\frac{1}{2}}{\frac{1}{2}})+c$
⇒ $\frac{2t^\frac{1}{2}}{49}(t+8)+c$
On replacing t with x terms
⇒ $\frac{2}{49}\sqrt{7x+9}(7x+9+8)+c$
⇒ $\frac{2}{49}(7x+17)(\sqrt{7x+9})+c$

Question 7. $\int\frac{x}{\sqrt{x+4}}dx$

Solution:

Given integral, $\int\frac{x}{\sqrt{x+4}}dx$
⇒ $\int\frac{x+4-4}{\sqrt{x+4}}dx$
⇒ $\int\frac{x+4}{\sqrt{x+4}}dx-4\int(x+4)^{\frac{-1}{2}}dx$
By using the formula,
$\int(ax+b)^ndx=\frac{(ax+b)^{n+1}}{a(n+1)}+c$ [where c is any arbitrary constant]
⇒ $\frac{(x+4)^{\frac{3}{2}}}{\frac{3}{2}}-\frac{4(x+4)^{\frac{1}{2}}}{\frac{1}{2}}+c$
⇒ $2(x+4)^{\frac{1}{2}}(\frac{x+4}{3}-4)+c$
⇒ $\frac{2}{3}(x-8){\sqrt{x+4}}+c$

Question 8. $\int\frac{2-3x}{\sqrt{1+3x}}dx$

Solution:

Given integral, $\int\frac{2-3x}{\sqrt{1+3x}}dx$
Let 1 + 3x = t
⇒ x = (t – 1)/3
On differentiating both sides, we get
dx = dt/3
On replacing x with t
⇒ $\int\frac{2-{\frac{3(t-1)}{3}}}{\sqrt{t}}\frac{dt}{3}$
⇒ $\frac{1}{3}\int\frac{3-t}{\sqrt{t}}dt$
⇒ $\frac{1}{3}\int3t^{\frac{-1}{2}}dt-\frac{1}{3}\int t^{\frac{1}{2}}dt$
By using the formula,
$\int x^{n}dx=\frac{x^{n+1}}{n+1}+c$ [where c is any arbitrary constant]
⇒ $\frac{1}{3}[\frac{3t^\frac{1}{2}}{\frac{1}{2}}]-\frac{1}{3}[\frac{t^\frac{3}{2}}{\frac{3}{2}}]+c$
Now on replacing t in terms of x
⇒ $2(1+3x)^{\frac{1}{2}}-\frac{2}{9}(1+3x)^{\frac{3}{2}}+c$
⇒ $2(1+3x)^{\frac{1}{2}}[1-\frac{1+3x}{9}]+c$
⇒ $\frac{2}{9}(8-3x)\sqrt{1+3x}+c$

Question 9. $\int(5x+3)\sqrt{2x-1}dx$

Solution:

Given integral, $\int(5x+3)\sqrt{2x-1}dx$
Let 2x – 1 = t²
⇒ x = (t²+ 1)/2
On differentiating on both sides,
dx = tdt
⇒ $\int(5(\frac{t^2+1}{2})+3)t^2dt$
⇒ $\frac{1}{2}\int(5t^2+11)t^2dt$
⇒ $\frac{1}{2}\int5t^4dt+\frac{1}{2}\int11t^2dt$
By using the formula,
$\int x^{n}dx=\frac{x^{n+1}}{n+1}+c$ [where c is any arbitrary constant]
⇒ $\frac{1}{2}{[\frac{5t^5}{5}]}+\frac{1}{2}{[\frac{11t^3}{3}]}+c$
On replacing t with x terms
⇒ $\frac{1}{2}[(2x-1)^{\frac{5}{2}}+\frac{11}{3}(2x-1)^{\frac{3}{2}}]+c$
⇒ $\frac{(2x-1)^{\frac{3}{2}}}{2}[2x-1+{\frac{11}{3}}]+c$
⇒ $\frac{(2x-1)^{\frac{3}{2}}}{2}(\frac{6x+8}{3})+c$
⇒ $\frac{1}{3}(3x+4)(2x-1)^{\frac{3}{2}}+c$

Question 10. $\int\frac{x}{{\sqrt{x+a}}-{\sqrt{x+b}}}dx$

Solution:

Given integral, $\int\frac{x}{{\sqrt{x+a}}-{\sqrt{x+b}}}dx$
On multiplying and dividing the given integral with $\sqrt{x+a}+\sqrt{x+b}$
We know that (a + b)(a – b) = a²– b²
⇒ $\int \frac{x(\sqrt{x+a}+{\sqrt{x+b}})}{(x+a)-(x+b)}dx$
⇒ $\int \frac{x\sqrt{x+a}+x\sqrt{x+b}}{a-b}dx$
⇒ $\frac{1}{a-b}[\int x\sqrt{x+a} dx+\int x\sqrt{x+b}dx]$
⇒ $\frac{1}{a-b}[\int(x+a){\sqrt{x+a}}dx-\int a{\sqrt{x+a}}dx+\int (x+b){\sqrt{x+b}}dx-\int{b\sqrt{x+b}}dx$
By using the formula,
$\int(ax+b)^ndx=\frac{(ax+b)^{n+1}}{a(n+1)}+c$ [where c is any arbitrary constant]
⇒ $\frac{1}{a-b}[\frac{(x+a)^\frac{5}{2}}{\frac{5}{2}}-{\frac{a(x+a)^{\frac{3}{2}}}{\frac{3}{2}}}+\frac{(x+b)^\frac{5}{2}}{\frac{5}{2}}-{\frac{b(x+b)^{\frac{3}{2}}}{\frac{3}{2}}}]+c$
⇒ $\frac{2}{a-b}[\frac{(x+a)^\frac{5}{2}+(x+b)^\frac{5}{2}}{5}-(\frac{a(x+a)^\frac{3}{2}+b(x+b)^\frac{3}{2}}{3})]+c$

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RD Sharma Class 12 Ex 19.5 Solutions Chapter 19 Indefinite Integrals

Question 1.

Question 2.

Question 3.

Question 4.

Question 5.

Question 6.

Question 7.

Question 8.

Question 9.

Question 10.

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Question 1. $\int\frac{x+1}{\sqrt{2x+3}}dx$

Question 2. $\int x\sqrt{x+2}dx$

Question 3. $\int\frac{x-1}{\sqrt{x+4}}dx$

Question 4. $\int(x+2)\sqrt{3x+5}dx$

Question 5. $\int\frac{2x+1}{\sqrt{3x+2}}dx$

Question 6. $\int\frac{3x+5}{\sqrt{7x+9}}dx$

Question 7. $\int\frac{x}{\sqrt{x+4}}dx$

Question 8. $\int\frac{2-3x}{\sqrt{1+3x}}dx$

Question 9. $\int(5x+3)\sqrt{2x-1}dx$

Question 10. $\int\frac{x}{{\sqrt{x+a}}-{\sqrt{x+b}}}dx$