Advertisements
Advertisements
प्रश्न
Evaluate the following:
`int ((x^2 + 2))/(x + 1) "d"x`
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"`
APPEARS IN
संबंधित प्रश्न
Evaluate the following definite integrals:
If \[f\left( a + b - x \right) = f\left( x \right)\] , then prove that \[\int_a^b xf\left( x \right)dx = \frac{a + b}{2} \int_a^b f\left( x \right)dx\]
If f (x) is a continuous function defined on [0, 2a]. Then, prove that
Evaluate each of the following integral:
Write the coefficient a, b, c of which the value of the integral
\[\int\limits_0^4 x\sqrt{4 - x} dx\]
\[\int\limits_1^2 x\sqrt{3x - 2} dx\]
\[\int\limits_0^\pi \sin^3 x\left( 1 + 2 \cos x \right) \left( 1 + \cos x \right)^2 dx\]
\[\int\limits_0^\infty \frac{x}{\left( 1 + x \right)\left( 1 + x^2 \right)} dx\]
\[\int\limits_{\pi/3}^{\pi/2} \frac{\sqrt{1 + \cos x}}{\left( 1 - \cos x \right)^{5/2}} dx\]
\[\int\limits_0^1 x \left( \tan^{- 1} x \right)^2 dx\]
\[\int\limits_0^\pi \frac{x \sin x}{1 + \cos^2 x} dx\]
\[\int\limits_{- 1}^1 e^{2x} dx\]
Evaluate the following:
Γ(4)
Evaluate the following integrals as the limit of the sum:
`int_0^1 x^2 "d"x`
Choose the correct alternative:
`int_(-1)^1 x^3 "e"^(x^4) "d"x` is
Find `int sqrt(10 - 4x + 4x^2) "d"x`
Evaluate `int (x^2"d"x)/(x^4 + x^2 - 2)`
`int (x + 3)/(x + 4)^2 "e"^x "d"x` = ______.
