हिंदी

Solve Each of the Following Integral: ∫ 4 2 X X 2 + 1 D X

Advertisements
Advertisements

प्रश्न

Solve each of the following integral:

\[\int_2^4 \frac{x}{x^2 + 1}dx\]
योग
Advertisements

उत्तर

\[\int_2^4 \frac{x}{x^2 + 1}dx\]
\[ = \frac{1}{2} \int_2^4 \frac{2x}{x^2 + 1}dx\]
\[ = \frac{1}{2} \times \left.\log\left( x^2 + 1 \right)\right|_2^4 ...................\left[ \int\frac{f'\left( x \right)}{f\left( x \right)}dx = \log f\left( x \right) + C \right]\]
\[ = \frac{1}{2}\left( \log17 - \log5 \right)\]
\[ = \frac{1}{2}\log\left( \frac{17}{5} \right) .............\left( \log a - \log b = \log\frac{a}{b} \right)\]

shaalaa.com
  क्या इस प्रश्न या उत्तर में कोई त्रुटि है?
अध्याय 19: Definite Integrals - Very Short Answers [पृष्ठ ११५]

APPEARS IN

आर.डी. शर्मा Mathematics Volume 1 and 2 [English] Class 12
अध्याय 19 Definite Integrals
Very Short Answers | Q 30 | पृष्ठ ११५

संबंधित प्रश्न

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

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

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

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

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

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

\[\int_0^\frac{\pi}{4} \left( \tan x + \cot x \right)^{- 2} dx\]

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

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

\[\int\limits_0^{\pi/2} \frac{dx}{a \cos x + b \sin x}a, b > 0\]

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

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

\[\int\limits_0^{\pi/2} \sin 2x \tan^{- 1} \left( \sin x \right) dx\]

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

\[\int_0^\frac{\pi}{2} \frac{\cos x}{\left( \cos\frac{x}{2} + \sin\frac{x}{2} \right)^n}dx\]

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

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

\[\int\limits_0^\pi \frac{x}{1 + \cos \alpha \sin x} dx, 0 < \alpha < \pi\]

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

If f(x) is a continuous function defined on [−aa], then prove that 

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

\[\int\limits_0^3 \left( x + 4 \right) dx\]

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

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

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

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

\[\int\limits_0^1 e^\left\{ x \right\} dx .\]

\[\int\limits_0^\infty \frac{1}{1 + e^x} dx\]  equals


Evaluate: \[\int\limits_{- \pi/2}^{\pi/2} \frac{\cos x}{1 + e^x}dx\] .

 

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


Using second fundamental theorem, evaluate the following:

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


Evaluate the following:

`int_0^2 "f"(x)  "d"x` where f(x) = `{{:(3 - 2x - x^2",", x ≤ 1),(x^2 + 2x - 3",", 1 < x ≤ 2):}`


Evaluate the following:

`Γ (9/2)`


Evaluate the following integrals as the limit of the sum:

`int_1^3 x  "d"x`


Choose the correct alternative:

If f(x) is a continuous function and a < c < b, then `int_"a"^"c" f(x)  "d"x + int_"c"^"b" f(x)  "d"x` is


Choose the correct alternative:

`int_0^oo x^4"e"^-x  "d"x` is


Evaluate `int (x^2"d"x)/(x^4 + x^2 - 2)`


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


`int x^3/(x + 1)` is equal to ______.


Given `int "e"^"x" (("x" - 1)/("x"^2)) "dx" = "e"^"x" "f"("x") + "c"`. Then f(x) satisfying the equation is:


Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×