English

∫ X √ X + a − √ X + B D X - Mathematics

Advertisements
Advertisements

Question

\[\int\frac{x}{\sqrt{x + a} - \sqrt{x + b}}dx\]
Sum
Advertisements

Solution

\[\int\frac{x}{\sqrt{x + a} - \sqrt{x + b}}dx\]
\[ = \int\frac{x}{\sqrt{x + a} - \sqrt{x + b}} \times \frac{\sqrt{x + a} + \sqrt{x + b}}{\sqrt{x + a} + \sqrt{x + b}}dx\]
\[ = \int\frac{x\left( \sqrt{x + a} + \sqrt{x + b} \right)}{\left( \sqrt{x + a} \right)^2 - \left( \sqrt{x + b} \right)^2}dx\]
\[ = ∫  \frac{x\left( \sqrt{x + a} + \sqrt{x + b} \right)}{x + a - x - b}dx\]
\[ = \frac{1}{a - b}\  ∫ x\left( \sqrt{x + a} + \sqrt{x + b} \right) dx\]
\[ = \frac{1}{a - b}\left[  ∫ x \left( \sqrt{x + a} \right) dx + \ ∫x\left( \sqrt{x + b} \right) dx \right]\]
\[ = \frac{1}{a - b}\left[ ∫ \left( x + a - a \right)\left( \sqrt{x + a} \right) dx + \int\left( x + b - b \right)\left( \sqrt{x + b} \right) dx \right]\]
\[ = \frac{1}{a - b}\left[ \int\left( x + a \right)\left( \sqrt{x + a} \right) dx - a\int\left( \sqrt{x + a} \right) dx + \int\left( x + b \right)\left( \sqrt{x + b} \right) dx - b\int\left( \sqrt{x + b} \right) dx \right]\]
\[ = \frac{1}{a - b}\left[ \int \left( x + a \right)^\frac{3}{2} dx - a\int \left( x + a \right)^\frac{1}{2} dx + \int \left( x + b \right)^\frac{3}{2} dx - b\int \left( x + b \right)^\frac{1}{2} dx \right]\]
\[ = \frac{1}{a - b}\left[ \frac{\left( x + a \right)^\frac{5}{2}}{\frac{5}{2}} - a\frac{\left( x + a \right)^\frac{3}{2}}{\frac{3}{2}} + \frac{\left( x + b \right)^\frac{5}{2}}{\frac{5}{2}} - b\frac{\left( x + b \right)^\frac{3}{2}}{\frac{3}{2}} \right] + \text{c             where, c is an arbitrary constant}\]
\[ = \frac{1}{a - b}\left[ \frac{2}{5} \left( x + a \right)^\frac{5}{2} - \frac{2a}{3} \left( x + a \right)^\frac{3}{2} + \frac{2}{5} \left( x + b \right)^\frac{5}{2} - \frac{2b}{3} \left( x + b \right)^\frac{3}{2} \right] + \text{c             where, c is an arbitrary constant}\]
\[Hence, \int\frac{x}{\sqrt{x + a} - \sqrt{x + b}}dx = \frac{1}{a - b}\left[ \frac{2}{5} \left( x + a \right)^\frac{5}{2} - \frac{2a}{3} \left( x + a \right)^\frac{3}{2} + \frac{2}{5} \left( x + b \right)^\frac{5}{2} - \frac{2b}{3} \left( x + b \right)^\frac{3}{2} \right] + \text{c           where, c is an arbitrary constant}\]

shaalaa.com
  Is there an error in this question or solution?
Chapter 19: Indefinite Integrals - Exercise 19.05 [Page 33]

APPEARS IN

RD Sharma Mathematics [English] Class 12
Chapter 19 Indefinite Integrals
Exercise 19.05 | Q 10 | Page 33

Video TutorialsVIEW ALL [1]

RELATED QUESTIONS

\[\int\sqrt{x}\left( 3 - 5x \right) dx\]

 


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

\[\int\frac{2 x^4 + 7 x^3 + 6 x^2}{x^2 + 2x} dx\]

If f' (x) = x − \[\frac{1}{x^2}\]  and  f (1)  \[\frac{1}{2},    find  f(x)\]

 


\[\int\frac{2 - 3x}{\sqrt{1 + 3x}} dx\]

\[\int\frac{\text{sin} \left( x - \alpha \right)}{\text{sin }\left( x + \alpha \right)} dx\]

\[\int\frac{e^{3x}}{e^{3x} + 1} dx\]

\[\int\frac{1}{x (3 + \log x)} dx\]

\[\int\frac{- \sin x + 2 \cos x}{2 \sin x + \cos x} dx\]

`  =  ∫ root (3){ cos^2 x}  sin x   dx `


\[\int \sin^5\text{ x }\text{cos x dx}\]

\[\int \tan^{3/2} x \sec^2 \text{x dx}\]

\[\int 5^{x + \tan^{- 1} x} . \left( \frac{x^2 + 2}{x^2 + 1} \right) dx\]

\[\int 5^{5^{5^x}} 5^{5^x} 5^x dx\]

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

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

\[\int \cot^5 \text{ x } {cosec}^4 x\text{ dx }\]

\[\int\frac{1}{x^2 - 10x + 34} dx\]

\[\int\frac{1}{\sqrt{\left( 1 - x^2 \right)\left\{ 9 + \left( \sin^{- 1} x \right)^2 \right\}}} dx\]

\[\int\frac{x}{\sqrt{x^2 + x + 1}} \text{ dx }\]

`int"x"^"n"."log"  "x"  "dx"`

\[\int \left( \log x \right)^2 \cdot x\ dx\]

\[\int {cosec}^3 x\ dx\]

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

\[\int \cos^{- 1} \left( 4 x^3 - 3x \right) \text{ dx }\]

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

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

\[\int\frac{5 x^2 - 1}{x \left( x - 1 \right) \left( x + 1 \right)} dx\]

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

Evaluate the following integral:

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

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

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

\[\int\frac{1}{1 - \cos x - \sin x} dx =\]

\[\int\sqrt{\frac{x}{1 - x}} dx\]  is equal to


\[\int\sin x \sin 2x \text{ sin  3x  dx }\]


\[\int\frac{1}{\text{ cos }\left( x - a \right) \text{ cos }\left( x - b \right)} \text{ dx }\]

\[\int\frac{1}{2 + \cos x} \text{ dx }\]


\[\int\frac{\sin^2 x}{\cos^6 x} \text{ dx }\]

Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×