हिंदी

Evaluate the following: d∫(x2+2)x+1dx

Advertisements
Advertisements

प्रश्न

Evaluate the following:

`int ((x^2 + 2))/(x + 1) "d"x`

योग
Advertisements

उत्तर

I = `int (x^2 + 2)/(x + 1) "d"x`

= `int (x^2 - 1 + 3)/(x + 1) "d"x`

= `int ((x - 1)(x + 1) + 3)/(x + 1) "d"x`

= `int (x - 1 + 3/(x + 1)) "d"x`

= `int (x - 1) "d"x + 3int 1/(x + 1) "d"x`

= `x^2/2 - x + 3 log |(x + 1)| + "C"`

shaalaa.com
  क्या इस प्रश्न या उत्तर में कोई त्रुटि है?
अध्याय 7: Integrals - Exercise [पृष्ठ १६३]

APPEARS IN

एनसीईआरटी एक्झांप्लर Mathematics Exemplar [English] Class 12
अध्याय 7 Integrals
Exercise | Q 3 | पृष्ठ १६३

संबंधित प्रश्न

\[\int\limits_4^9 \frac{1}{\sqrt{x}} dx\]

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

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

\[\int\limits_0^{\pi/2} \frac{1}{5 \cos x + 3 \sin x} dx\]

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

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

\[\int_\frac{1}{3}^1 \frac{\left( x - x^3 \right)^\frac{1}{3}}{x^4}dx\]

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

Evaluate the following integral:

\[\int\limits_{- 3}^3 \left| x + 1 \right| dx\]

\[\int_{- 1}^2 \left( \left| x + 1 \right| + \left| x \right| + \left| x - 1 \right| \right)dx\]

 


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

Evaluate each of the following integral:

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


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

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

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

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

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

\[\int\limits_1^3 \left( 2 x^2 + 5x \right) dx\]

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

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

Evaluate each of the following  integral:

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

 


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

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


\[\int_0^\frac{\pi^2}{4} \frac{\sin\sqrt{x}}{\sqrt{x}} dx\] equals


\[\int\limits_1^e \log x\ dx =\]

\[\int\limits_1^\sqrt{3} \frac{1}{1 + x^2} dx\]  is equal to ______.

\[\int\limits_{\pi/6}^{\pi/3} \frac{1}{\sin 2x} dx\]  is equal to

\[\lim_{n \to \infty} \left\{ \frac{1}{2n + 1} + \frac{1}{2n + 2} + . . . + \frac{1}{2n + n} \right\}\] is equal to

The value of \[\int\limits_0^{\pi/2} \log\left( \frac{4 + 3 \sin x}{4 + 3 \cos x} \right) dx\] is 

 


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

 

\[\int\limits_0^\pi x \sin x \cos^4 x dx\]


\[\int\limits_0^\pi \frac{x \tan x}{\sec x + \tan x} dx\]


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


Evaluate the following using properties of definite integral:

`int_(- pi/4)^(pi/4) x^3 cos^3 x  "d"x`


Evaluate the following:

`Γ (9/2)`


Evaluate the following:

`int_0^oo "e"^(-mx) x^6 "d"x`


Choose the correct alternative:

`int_0^1 (2x + 1)  "d"x` is


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


Evaluate `int "dx"/sqrt((x - alpha)(beta - x)), beta > alpha`


Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×