हिंदी

2 ∫ 1 X + 3 X ( X + 2 ) D X

Advertisements
Advertisements

प्रश्न

\[\int\limits_1^2 \frac{x + 3}{x \left( x + 2 \right)} dx\]
Advertisements

उत्तर

\[Let\ I = \int_1^2 \frac{x + 3}{x\left( x + 2 \right)} d x . Then, \]
\[I = \int_1^2 \left( \frac{x}{x\left( x + 2 \right)} + \frac{3}{x\left( x + 2 \right)} \right) d x\]
\[ \Rightarrow I = \int_1^2 \frac{dx}{\left( x + 2 \right)} + \int_1^2 \frac{3}{x\left( x + 2 \right)} d x\]
\[ \Rightarrow I = \left[ \log \left( x + 2 \right) \right]_1^2 + \frac{3}{2} \int_1^2 \left( \frac{1}{x} - \frac{1}{x + 2} \right) dx\]
\[ \Rightarrow I = \left[ \log \left( x + 2 \right) \right]_1^2 + \frac{3}{2} \left[ \log x - \log \left( x + 2 \right) \right]_1^2 \]
\[ \Rightarrow I = \log 4 - \log 3 + \frac{3}{2}\left[ \log 2 - \log 4 - 0 + \log 3 \right]\]
\[ \Rightarrow I = \log 4 - \log 3 + \frac{3}{2}\left[ - \log 2 + \log 3 \right]\]
\[ \Rightarrow I = 2 \log 2 - \log 3 + \frac{3}{2} \log 3 - \frac{3}{2} \log 2\]
\[ \Rightarrow I = \frac{1}{2} \log 2 + \frac{1}{2} \log 3\]
\[ \Rightarrow I = \frac{1}{2}\left( \log 2 + \log 3 \right)\]
\[ \Rightarrow I = \frac{1}{2} \log 6\]
shaalaa.com
  क्या इस प्रश्न या उत्तर में कोई त्रुटि है?
अध्याय 19: Definite Integrals - Exercise 20.1 [पृष्ठ १७]

APPEARS IN

आर.डी. शर्मा Mathematics Volume 1 and 2 [English] Class 12
अध्याय 19 Definite Integrals
Exercise 20.1 | Q 37 | पृष्ठ १७

संबंधित प्रश्न

\[\int\limits_0^1 \frac{1}{1 + x^2} dx\]

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

\[\int\limits_0^{\pi/2} \cos^4\ x\ dx\]

 


\[\int\limits_0^{\pi/2} x^2 \cos\ 2x\ dx\]

\[\int\limits_0^1 x \left( 1 - x \right)^5 dx\]

\[\int\limits_0^{\pi/2} \sqrt{\sin \phi} \cos^5 \phi\ d\phi\]

 


\[\int\limits_0^{\pi/2} \frac{\cos x}{1 + \sin^2 x} dx\]

\[\int\limits_0^{\pi/3} \frac{\cos x}{3 + 4 \sin x} dx\]

\[\int_0^\frac{1}{2} \frac{x \sin^{- 1} x}{\sqrt{1 - x^2}}dx\]

\[\int\limits_0^1 \frac{\tan^{- 1} x}{1 + x^2} dx\]

\[\int\limits_0^1 \frac{1 - x^2}{\left( 1 + x^2 \right)^2} dx\]

\[\int\limits_0^a x \sqrt{\frac{a^2 - x^2}{a^2 + x^2}} dx\]

\[\int_{- 2}^2 x e^\left| x \right| dx\]

\[\int_{- \frac{\pi}{2}}^\pi \sin^{- 1} \left( \sin x \right)dx\]

\[\int_0^{2\pi} \cos^{- 1} \left( \cos x \right)dx\]

\[\int\limits_0^{\pi/2} \left( 2 \log \cos x - \log \sin 2x \right) dx\]

 


\[\int\limits_0^\pi \log\left( 1 - \cos x \right) dx\]

Evaluate the following integral:

\[\int_{- 1}^1 \left| xcos\pi x \right|dx\]

 


\[\int\limits_0^2 \left( x^2 + 4 \right) dx\]

\[\int\limits_{- 2}^1 \frac{\left| x \right|}{x} dx .\]

\[\int\limits_0^1 \frac{1}{1 + x^2} dx\]

Write the coefficient abc of which the value of the integral

\[\int\limits_{- 3}^3 \left( a x^2 + bx + c \right) dx\] is independent.

Evaluate : 

\[\int\limits_2^3 3^x dx .\]

\[\int\limits_0^1 2^{x - \left[ x \right]} dx\]

If \[\left[ \cdot \right] and \left\{ \cdot \right\}\] denote respectively the greatest integer and fractional part functions respectively, evaluate the following integrals:

\[\int\limits_0^{\pi/4} \sin \left\{ x \right\} dx\]

 


\[\int\limits_{- 1}^1 \left| 1 - x \right| dx\]  is equal to

If \[I_{10} = \int\limits_0^{\pi/2} x^{10} \sin x\ dx,\]  then the value of I10 + 90I8 is

 


The value of \[\int\limits_{- \pi/2}^{\pi/2} \left( x^3 + x \cos x + \tan^5 x + 1 \right) dx, \] is 


\[\int\limits_1^2 x\sqrt{3x - 2} dx\]


\[\int\limits_0^{\pi/2} \frac{\sin x}{\sqrt{1 + \cos x}} dx\]


\[\int\limits_0^{\pi/2} \frac{\cos x}{1 + \sin^2 x} dx\]


\[\int\limits_0^{\pi/4} \sin 2x \sin 3x dx\]


\[\int\limits_{- \pi/4}^{\pi/4} \left| \tan x \right| dx\]


\[\int\limits_2^3 \frac{\sqrt{x}}{\sqrt{5 - x} + \sqrt{x}} dx\]


Choose the correct alternative:

The value of `int_(- pi/2)^(pi/2) cos  x  "d"x` is


Find `int x^2/(x^4 + 3x^2 + 2) "d"x`


`int "e"^x ((1 - x)/(1 + x^2))^2  "d"x` is equal to ______.


Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×