मराठी

∫ 1 0 X Log ( 1 + 2 X ) D X

Advertisements
Advertisements

प्रश्न

\[\int_0^1 x\log\left( 1 + 2x \right)dx\]
बेरीज
Advertisements

उत्तर

\[\text{Let I }=\int_0^1 x\log\left( 1 + 2x \right)dx\]

Applying integration by parts, we have

\[I = \log\left( 1 + 2x \right)\frac{x^2}{2}_0^1 - \int_0^1 \left( \frac{2}{1 + 2x} \right) \times \frac{x^2}{2}dx\]
\[ = \frac{1}{2}\left( \log3 - 0 \right) - \int_0^1 \frac{x^2}{1 + 2x}dx\]
\[ = \frac{1}{2}\log3 - \frac{1}{4} \int_0^1 \frac{4 x^2 - 1 + 1}{1 + 2x}dx\]
\[ = \frac{1}{2}\log3 - \frac{1}{4} \int_0^1 \frac{\left( 2x + 1 \right)\left( 2x - 1 \right)}{1 + 2x}dx - \frac{1}{4} \int_0^1 \frac{1}{1 + 2x}dx\]
\[ = \frac{1}{2}\log3 - \frac{1}{4} \int_0^1 \left( 2x - 1 \right)dx - \frac{1}{4} \int_0^1 \frac{1}{1 + 2x}dx\]

\[= \left.\frac{1}{2}\log3 - \frac{1}{4} \times \frac{\left( 2x - 1 \right)^2}{2 \times 2}\right|_0^1 - \left.\frac{1}{4} \times \frac{\log\left( 1 + 2x \right)}{2}\right|_0^1 \]
\[ = \frac{1}{2}\log3 - \frac{1}{16}\left( 1 - 1 \right) - \frac{1}{8}\left( \log3 - \log1 \right)\]
\[ = \frac{1}{2}\log3 - 0 - \frac{1}{8}\log3 ....................\left( \log1 = 0 \right)\]
\[ = \frac{3}{8}\log3\]

 

shaalaa.com
  या प्रश्नात किंवा उत्तरात काही त्रुटी आहे का?
पाठ 19: Definite Integrals - Exercise 20.1 [पृष्ठ १८]

APPEARS IN

आर.डी. शर्मा Mathematics Volume 1 and 2 [English] Class 12
पाठ 19 Definite Integrals
Exercise 20.1 | Q 65 | पृष्ठ १८

संबंधित प्रश्‍न

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

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

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

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

\[\int\limits_{- 1}^1 \frac{1}{x^2 + 2x + 5} dx\]

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

\[\int\limits_0^{2\pi} e^x \cos\left( \frac{\pi}{4} + \frac{x}{2} \right) dx\]

\[\int\limits_1^2 \frac{3x}{9 x^2 - 1} dx\]

\[\int\limits_0^{\pi/2} \frac{\sin x \cos x}{1 + \sin^4 x} dx\]

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

\[\int\limits_0^{( \pi )^{2/3}} \sqrt{x} \cos^2 x^{3/2} dx\]


\[\int\limits_0^\pi \sin^3 x\left( 1 + 2 \cos x \right) \left( 1 + \cos x \right)^2 dx\]

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

Evaluate each of the following integral:

\[\int_a^b \frac{x^\frac{1}{n}}{x^\frac{1}{n} + \left( a + b - x \right)^\frac{1}{n}}dx, n \in N, n \geq 2\]


\[\int\limits_0^5 \frac{\sqrt[4]{x + 4}}{\sqrt[4]{x + 4} + \sqrt[4]{9 - x}} dx\]

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

 


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

\[\int\limits_0^\pi x \cos^2 x\ dx\]

If f (x) is a continuous function defined on [0, 2a]. Then, prove that

\[\int\limits_0^{2a} f\left( x \right) dx = \int\limits_0^a \left\{ f\left( x \right) + f\left( 2a - x \right) \right\} dx\]

 


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

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

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

\[\int\limits_0^\pi \cos^5 x\ dx .\]

\[\int\limits_{- \pi/2}^{\pi/2} \log\left( \frac{a - \sin \theta}{a + \sin \theta} \right) d\theta\]

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

Given that \[\int\limits_0^\infty \frac{x^2}{\left( x^2 + a^2 \right)\left( x^2 + b^2 \right)\left( x^2 + c^2 \right)} dx = \frac{\pi}{2\left( a + b \right)\left( b + c \right)\left( c + a \right)},\] the value of \[\int\limits_0^\infty \frac{dx}{\left( x^2 + 4 \right)\left( x^2 + 9 \right)},\]


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

Evaluate : \[\int e^{2x} \cdot \sin \left( 3x + 1 \right) dx\] .


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


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


\[\int\limits_0^\pi \frac{dx}{6 - \cos x}dx\]


Find : `∫_a^b logx/x` dx


Evaluate the following using properties of definite integral:

`int_(- pi/2)^(pi/2) sin^2theta  "d"theta`


Evaluate the following:

`Γ (9/2)`


Evaluate the following integrals as the limit of the sum:

`int_0^1 (x + 4)  "d"x`


Evaluate the following integrals as the limit of the sum:

`int_1^3 (2x + 3)  "d"x`


Choose the correct alternative:

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


`int x^9/(4x^2 + 1)^6  "d"x` is equal to ______.


Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×