English

CBSECommerce (English Medium) Class 12

Question Papers

Question Papers3000

Textbook Solutions35049

MCQ Online Mock Tests42

Important Solutions23871

Concept Notes & Videos707

∫ 1 0 X Log ( 1 + 2 X ) D X

Advertisements

Advertisements

Question

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

Sum

Advertisements

Solution

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

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

Definite Integrals

Is there an error in this question or solution?

Chapter 19: Definite Integrals - Exercise 20.1 [Page 18]

APPEARS IN

R.D. Sharma Mathematics Volume 1 and 2 [English] Class 12

Chapter 19 Definite Integrals
Exercise 20.1 | Q 65 | Page 18

RELATED QUESTIONS

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

\[\int\limits_0^\infty e^{- x} dx\]

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

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

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

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

\[\int_0^\frac{\pi}{4} \frac{\sin x + \cos x}{3 + \sin2x}dx\]

\[\int\limits_0^{\pi/4} \sin^3 2t \cos 2t\ dt\]

\[\int\limits_0^{\pi/2} \cos^5 x\ dx\]

\[\int_0^\frac{1}{2} \frac{1}{\left( 1 + x^2 \right)\sqrt{1 - x^2}}dx\]

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

\[\int\limits_0^\pi x \sin^3 x\ dx\]

\[\int\limits_{- \pi/2}^{\pi/2} \sin^3 x\ dx\]

\[\int\limits_3^5 \left( 2 - x \right) dx\]

\[\int\limits_0^2 e^x dx\]

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

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

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

\[\int\limits_{- \pi/2}^{\pi/2} \cos^2 x\ dx .\]

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

\[\int\limits_a^b \frac{f\left( x \right)}{f\left( x \right) + f\left( a + b - x \right)} dx .\]

The value of \[\int\limits_0^{2\pi} \sqrt{1 + \sin\frac{x}{2}}dx\] is

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

The value of the integral \[\int\limits_0^\infty \frac{x}{\left( 1 + x \right)\left( 1 + x^2 \right)} dx\]

\[\int\limits_0^1 \frac{d}{dx}\left\{ \sin^{- 1} \left( \frac{2x}{1 + x^2} \right) \right\} dx\] is equal to

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

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

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

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

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

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

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

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

\[\int\limits_0^4 x dx\]

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

Using second fundamental theorem, evaluate the following:

`int_0^1 "e"^(2x) "d"x`

Evaluate the following:

f(x) = `{{:("c"x",", 0 < x < 1),(0",", "otherwise"):}` Find 'c" if `int_0^1 "f"(x) "d"x` = 2

Evaluate the following using properties of definite integral:

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

Evaluate the following integrals as the limit of the sum:

`int_1^3 (2x + 3) "d"x`

Find `int sqrt(10 - 4x + 4x^2) "d"x`

Notifications