RD Sharma Class 12 Ex 19.5 Solutions Chapter 19 Indefinite Integrals

Here we provide RD Sharma Class 12 Ex 19.5 Solutions Chapter 19 Indefinite Integrals for English medium students, Which will very helpful for every student in their exams. Students can download the latest Sharma Class 12 Ex 19.5 Solutions Chapter 19 Indefinite Integrals book pdf download. Now you will get step-by-step solutions to each question.

TextbookNCERT
ClassClass 12th
SubjectMaths
Chapter19
Exercise19.5
CategoryRD Sharma Solutions

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 = (t+ 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) = a– 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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