Advertisements
Advertisements
प्रश्न
Find `int sqrt(10 - 4x + 4x^2) "d"x`
उत्तर
We have I = `int sqrt(10 - 4x + 4x^2) "d"x`
= `int sqrt((2x - 1)^2 + (3)^2) "d"x`
Put t = 2x – 1
Then dt = 2dx.
Therefore, I = `1/2 int sqrt("t"^2 + (3)^2) "dt"`
= `1/2 "t" sqrt("t"^2 + 9)/2 + 9/4 log|"t" + sqrt("t"^2 + 9)| + "C"`
= `1/4(2x - 1) sqrt((2x - 1)^2 + 9) + 9/4 log|(2x - 1) + sqrt((2x - 1)^2 + 9)| + "C"`
APPEARS IN
संबंधित प्रश्न
\[\int\limits_{\pi/4}^{\pi/2} \cot x\ dx\]
Evaluate the following definite integrals:
Evaluate the following integral:
If f(2a − x) = −f(x), prove that
If \[\int\limits_0^a 3 x^2 dx = 8,\] write the value of a.
If \[\left[ \cdot \right] and \left\{ \cdot \right\}\] denote respectively the greatest integer and fractional part functions respectively, evaluate the following integrals:
The value of \[\int\limits_0^{2\pi} \sqrt{1 + \sin\frac{x}{2}}dx\] is
`int_0^1 sqrt((1 - "x")/(1 + "x")) "dx"`
The value of \[\int\limits_0^1 \tan^{- 1} \left( \frac{2x - 1}{1 + x - x^2} \right) dx,\] is
\[\int\limits_1^2 x\sqrt{3x - 2} dx\]
\[\int\limits_0^1 \tan^{- 1} x dx\]
\[\int\limits_0^1 \tan^{- 1} \left( \frac{2x}{1 - x^2} \right) dx\]
\[\int\limits_0^\pi \sin^3 x\left( 1 + 2 \cos x \right) \left( 1 + \cos x \right)^2 dx\]
\[\int\limits_1^2 \frac{x + 3}{x\left( x + 2 \right)} dx\]
\[\int\limits_0^1 \left| \sin 2\pi x \right| dx\]
\[\int\limits_0^\pi \frac{x \tan x}{\sec x + \tan x} dx\]
Evaluate the following:
`int ((x^2 + 2))/(x + 1) "d"x`
