English

∫ 1 √ X ( √ X + 1 ) D X - Mathematics

Advertisements
Advertisements

Question

\[\int\frac{1}{\sqrt{x}\left( \sqrt{x} + 1 \right)} dx\]
Sum
Advertisements

Solution

\[\text{Let I} = \int\frac{1}{\sqrt{x}\left( \sqrt{x} + 1 \right)}dx\]
\[\text{Putting}\ \sqrt{x} + 1 = t\]
\[ \Rightarrow \frac{1}{2\sqrt{x}} = \frac{dt}{dx}\]
\[ \Rightarrow \frac{1}{\sqrt{x}}dx = 2dt\]
\[ \therefore I = 2\int\frac{1}{t}dt\]
\[ =\text{ 2 }\text{ln}\left| t \right| + C\]
\[ = \text{2 }\text{ln} \left| \sqrt{x} + 1 \right| + C \left[ \because t = \sqrt{x} + 1 \right]\]

shaalaa.com
  Is there an error in this question or solution?
Chapter 19: Indefinite Integrals - Exercise 19.08 [Page 48]

APPEARS IN

RD Sharma Mathematics [English] Class 12
Chapter 19 Indefinite Integrals
Exercise 19.08 | Q 45 | Page 48

RELATED QUESTIONS

Integrate the following w.r.t. x `(x^3-3x+1)/sqrt(1-x^2)`


\[\int\frac{x - 1}{\sqrt{x + 4}} dx\]

\[\int\frac{x}{\sqrt{x + 4}} dx\]

` ∫ {cot x}/ { log sin x} dx `

\[\int\frac{e^{2x}}{e^{2x} - 2} dx\]

` ∫  {1+tan}/{ x + log  sec  x   dx} `

\[\int\frac{e^{x - 1} + x^{e - 1}}{e^x + x^e} dx\]

\[\int\frac{1}{\sin x \cos^2 x} dx\]

\[\int\frac{e^x}{\left( 1 + e^x \right)^2} dx\]

\[\int\frac{\cot x}{\sqrt{\sin x}} dx\]


 `   ∫     tan x    .  sec^2 x   \sqrt{1 - tan^2 x}     dx\ `

Evaluate the following integrals: 

\[\int\frac{x + 2}{\sqrt{x^2 + 2x + 3}} \text{ dx }\]

Evaluate the following integrals:

\[\int\frac{x \cos^{- 1} x}{\sqrt{1 - x^2}}dx\]

 


Evaluate the following integrals:

\[\int\frac{\log x}{\left( x + 1 \right)^2}dx\]

 


Evaluate the following integrals:

\[\int e^{2x} \left( \frac{1 - \sin2x}{1 - \cos2x} \right)dx\]

Evaluate the following integrals:

\[\int\left( x + 3 \right)\sqrt{3 - 4x - x^2} \text{  dx }\]

\[\int(3x + 1) \sqrt{4 - 3x - 2 x^2} \text{  dx }\]

\[\int\frac{a x^2 + bx + c}{\left( x - a \right) \left( x - b \right) \left( x - c \right)} dx,\text{ where a, b, c are distinct}\]

Evaluate the following integral :-

\[\int\frac{x}{\left( x^2 + 1 \right)\left( x - 1 \right)}dx\]

Evaluate the following integral:

\[\int\frac{x^2}{\left( x^2 + 4 \right)\left( x^2 + 9 \right)}dx\]

Evaluate the following integral:

\[\int\frac{3x - 2}{\left( x + 1 \right)^2 \left( x + 3 \right)}dx\]

Evaluate the following integral:

\[\int\frac{2 x^2 + 1}{x^2 \left( x^2 + 4 \right)}dx\]

Evaluate the following integrals:

\[\int\frac{x^2}{(x - 1) ( x^2 + 1)}dx\]

Evaluate:

\[\int\frac{x^2 + 4x}{x^3 + 6 x^2 + 5} \text{ dx }\]

Evaluate:\[\int\frac{\sin \sqrt{x}}{\sqrt{x}} \text{ dx }\]


Evaluate:\[\int\frac{\left( 1 + \log x \right)^2}{x} \text{   dx }\]


Evaluate:\[\int \sec^2 \left( 7 - 4x \right) \text{ dx }\]


Write the value of\[\int\sec x \left( \sec x + \tan x \right)\text{  dx }\]


Evaluate: \[\int\frac{1}{x^2 + 16}\text{ dx }\]


Evaluate:  \[\int\frac{2}{1 - \cos2x}\text{ dx }\]


Evaluate : \[\int\frac{1}{x(1 + \log x)} \text{ dx}\]


Evaluate: `int_  (x + sin x)/(1 + cos x )  dx`


Evaluate the following:

`int ("d"x)/sqrt(16 - 9x^2)`


Evaluate the following:

`int (3x - 1)/sqrt(x^2 + 9) "d"x`


Evaluate the following:

`int sqrt(2"a"x - x^2)  "d"x`


Evaluate the following:

`int_1^2 ("d"x)/sqrt((x - 1)(2 - x))`


Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×