हिंदी

Evaluate the following: d∫(x2+2)x+1dx - Mathematics

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

APPEARS IN

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

संबंधित प्रश्न

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

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

\[\int\limits_0^{\pi/6} \cos x \cos 2x\ dx\]

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

\[\int_0^{2\pi} \sqrt{1 + \sin\frac{x}{2}}dx\]

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

\[\int\limits_0^{\pi/6} \cos^{- 3} 2 \theta \sin 2\ \theta\ d\ \theta\]

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

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

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

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

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

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

Evaluate the following integral:

\[\int_{- a}^a \log\left( \frac{a - \sin\theta}{a + \sin\theta} \right)d\theta\]

If `f` is an integrable function such that f(2a − x) = f(x), then prove that

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

 


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

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

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

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

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

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

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

Evaluate each of the following integral:

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

 


Solve each of the following integral:

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

Evaluate : 

\[\int\limits_2^3 3^x dx .\]

\[\int\limits_0^1 \left\{ x \right\} dx,\] where {x} denotes the fractional part of x.  

 

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


The value of \[\int\limits_0^{\pi/2} \cos x\ e^{\sin x}\ dx\] is

 


Evaluate : \[\int\limits_0^{2\pi} \cos^5 x dx\] .


`int_0^(2a)f(x)dx`


\[\int\limits_1^5 \frac{x}{\sqrt{2x - 1}} dx\]


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


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


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


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


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


Find : `∫_a^b logx/x` 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)`


Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×