मराठी

2 ∫ 1 X ( X + 1 ) ( X + 2 ) D X - Mathematics

Advertisements
Advertisements

प्रश्न

\[\int\limits_1^2 \frac{x}{\left( x + 1 \right) \left( x + 2 \right)} dx\]
Advertisements

उत्तर

\[Let\ I = \int_1^2 \frac{x}{\left( x + 1 \right)\left( x + 2 \right)} d\ x\ . Then, \]
\[I = \int_1^2 \left( \frac{- 1}{\left( x + 1 \right)} + \frac{2}{\left( x + 2 \right)} \right) d x\]
\[ \Rightarrow I = - \int_1^2 \frac{1}{\left( x + 1 \right)} dx + 2 \int_1^2 \frac{1}{\left( x + 2 \right)} dx\]
\[ \Rightarrow I = \left[ - \log \left( x + 1 \right) + 2 \log \left( x + 2 \right) \right]_1^2 \]
\[ \Rightarrow I = - \log 3 + 2 \log 4 + \log 2 - 2 \log 3\]
\[ \Rightarrow I = 5 \log 2 - 3 \log 3\]
\[ \Rightarrow I = \log 2^5 - \log 3^3 \]
\[ \Rightarrow I = \log \frac{32}{27}\]

shaalaa.com
Definite Integrals
  या प्रश्नात किंवा उत्तरात काही त्रुटी आहे का?
पाठ 20: Definite Integrals - Exercise 20.1 [पृष्ठ १७]

APPEARS IN

आरडी शर्मा Mathematics [English] Class 12
पाठ 20 Definite Integrals
Exercise 20.1 | Q 55 | पृष्ठ १७

संबंधित प्रश्‍न

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

\[\int\limits_1^3 \frac{\log x}{\left( x + 1 \right)^2} dx\]

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

\[\int\limits_0^{\pi/4} \left( \sqrt{\tan}x + \sqrt{\cot}x \right) dx\]

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

\[\int\limits_1^4 f\left( x \right) dx, where f\left( x \right) = \begin{cases}7x + 3 & , & \text{if }1 \leq x \leq 3 \\ 8x & , & \text{if }3 \leq x \leq 4\end{cases}\]


\[\int_{- \frac{\pi}{2}}^\frac{\pi}{2} \frac{- \frac{\pi}{2}}{\sqrt{\cos x \sin^2 x}}dx\]

If  \[f\left( a + b - x \right) = f\left( x \right)\] , then prove that \[\int_a^b xf\left( x \right)dx = \frac{a + b}{2} \int_a^b f\left( x \right)dx\]

 


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

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

 


\[\int\limits_1^4 \left( x^2 - x \right) dx\]

\[\int\limits_0^3 \left( 2 x^2 + 3x + 5 \right) dx\]

\[\int\limits_a^b x\ dx\]

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

\[\int\limits_0^3 \frac{1}{x^2 + 9} dx .\]

\[\int\limits_0^2 \sqrt{4 - x^2} dx\]

Evaluate each of the following  integral:

\[\int_0^1 x e^{x^2} dx\]

 


\[\int\limits_0^2 \left[ x \right] dx .\]

`int_0^1 sqrt((1 - "x")/(1 + "x")) "dx"`


\[\int\limits_{\pi/6}^{\pi/3} \frac{1}{1 + \sqrt{\cot}x} dx\] is

\[\int\limits_0^{\pi/2} \frac{1}{1 + \tan x} dx\]  is equal to

If \[\int\limits_0^1 f\left( x \right) dx = 1, \int\limits_0^1 xf\left( x \right) dx = a, \int\limits_0^1 x^2 f\left( x \right) dx = a^2 , then \int\limits_0^1 \left( a - x \right)^2 f\left( x \right) dx\] equals


\[\int\limits_0^{\pi/2} x \sin x\ dx\]  is equal to

`int_0^(2a)f(x)dx`


\[\int\limits_0^4 x\sqrt{4 - x} dx\]


\[\int\limits_0^{1/\sqrt{3}} \tan^{- 1} \left( \frac{3x - x^3}{1 - 3 x^2} \right) dx\]


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


\[\int\limits_{\pi/3}^{\pi/2} \frac{\sqrt{1 + \cos x}}{\left( 1 - \cos x \right)^{5/2}} dx\]


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


Using second fundamental theorem, evaluate the following:

`int_1^"e" ("d"x)/(x(1 + logx)^3`


Evaluate the following using properties of definite integral:

`int_0^1 log (1/x - 1)  "d"x`


Evaluate the following using properties of definite integral:

`int_0^1 x/((1 - x)^(3/4))  "d"x`


Evaluate the following:

`Γ (9/2)`


Choose the correct alternative:

`int_0^1 (2x + 1)  "d"x` is


Choose the correct alternative:

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


Choose the correct alternative:

`Γ(3/2)`


Verify the following:

`int (2x + 3)/(x^2 + 3x) "d"x = log|x^2 + 3x| + "C"`


If `intx^3/sqrt(1 + x^2) "d"x = "a"(1 + x^2)^(3/2) + "b"sqrt(1 + x^2) + "C"`, then ______.


Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×