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Class 12 RD Sharma Solutions – Chapter 19 Indefinite Integrals – Exercise 19.5

  • Last Updated : 07 Apr, 2021

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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