English

CBSECommerce (English Medium) Class 12

Question Papers

Question Papers2580

Textbook Solutions20345

MCQ Online Mock Tests42

Important Solutions22654

Concept Notes & Videos608

∫ 1 0 X Log ( 1 + 2 X ) D X - Mathematics

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 20: Definite Integrals - Exercise 20.1 [Page 18]

APPEARS IN

RD Sharma Mathematics [English] Class 12

Chapter 20 Definite Integrals
Exercise 20.1 | Q 65 | Page 18

RELATED QUESTIONS

\[\int\limits_{- 2}^3 \frac{1}{x + 7} dx\]

\[\int\limits_2^3 \frac{x}{x^2 + 1} dx\]

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

\[\int\limits_{- \pi/4}^{\pi/4} \frac{1}{1 + \sin x} dx\]

\[\int\limits_{\pi/2}^\pi e^x \left( \frac{1 - \sin x}{1 - \cos x} \right) dx\]

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

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

\[\int\limits_2^4 \frac{x}{x^2 + 1} dx\]

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

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

\[\int\limits_0^9 f\left( x \right) dx, where f\left( x \right) \begin{cases}\sin x & , & 0 \leq x \leq \pi/2 \\ 1 & , & \pi/2 \leq x \leq 3 \\ e^{x - 3} & , & 3 \leq x \leq 9\end{cases}\]

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

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

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

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

\[\int\limits_{- \pi/4}^{\pi/4} \sin^2 x\ dx\]

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

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

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

\[\int\limits_0^{\pi/2} \frac{\sin^n x}{\sin^n x + \cos^n x} dx, n \in N .\]

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

Evaluate each of the following integral:

\[\int_0^\frac{\pi}{4} \tan\ xdx\]

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

Evaluate each of the following integral:

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

If \[\int_0^a \frac{1}{4 + x^2}dx = \frac{\pi}{8}\] , find the value of a.

The value of \[\int\limits_0^\pi \frac{x \tan x}{\sec x + \cos x} dx\] is __________ .

\[\int\limits_0^{\pi/2} \frac{\cos x}{\left( 2 + \sin x \right)\left( 1 + \sin x \right)} dx\] equals

\[\int\limits_{- \pi/2}^{\pi/2} \sin\left| x \right| dx\] is equal to

\[\int\limits_0^{\pi/2} \frac{1}{1 + \cot^3 x} dx\] is equal to

Evaluate : \[\int\frac{dx}{\sin^2 x \cos^2 x}\] .

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

\[\int\limits_0^1 \log\left( 1 + x \right) dx\]

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

\[\int\limits_{\pi/6}^{\pi/2} \frac{\ cosec x \cot x}{1 + {cosec}^2 x} dx\]

Using second fundamental theorem, evaluate the following:

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

Using second fundamental theorem, evaluate the following:

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

Using second fundamental theorem, evaluate the following:

`int_1^2 (x - 1)/x^2 "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_0^(i/2) (sin^7x)/(sin^7x + cos^7x) "d"x`

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

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

Notifications