Advertisements
Advertisements
प्रश्न
`int 1/sqrt(x^2 - 8x - 20) "d"x`
उत्तर
Let I = `int 1/sqrt(x^2 - 8x - 20) "d"x`
= `int 1/sqrt(x^2 - 2.4x + 16 - 16 - 20) "d"x`
= `int ("d"x)/sqrt((x - 4)^2 - 36) "d"x`
= `int ("d"x)/sqrt((x - 4)^2 - 6^2) "d"x`
= `log|(x - 4) + sqrt((x - 4)^2 - 6^2)| + "c"`
∴ I = `log|(x - 4) + sqrt(x^2 - 8x - 20)| + "c"`
APPEARS IN
संबंधित प्रश्न
Find :
`∫(log x)^2 dx`
Evaluate the following : `int x^2.log x.dx`
Evaluate the following : `int x^3.tan^-1x.dx`
Evaluate the following : `int x^3.logx.dx`
Evaluate the following:
`int x.sin 2x. cos 5x.dx`
Integrate the following functions w.r.t. x : `x^2 .sqrt(a^2 - x^6)`
Integrate the following functions w.r.t. x : `sqrt(4^x(4^x + 4))`
Integrate the following functions w.r.t. x : `sqrt(x^2 + 2x + 5)`
Integrate the following w.r.t.x : `log (1 + cosx) - xtan(x/2)`
Integrate the following w.r.t.x : e2x sin x cos x
Evaluate: `int "dx"/("9x"^2 - 25)`
`int (sin(x - "a"))/(cos (x + "b")) "d"x`
`int (cos2x)/(sin^2x cos^2x) "d"x`
The value of `int "e"^(5x) (1/x - 1/(5x^2)) "d"x` is ______.
Evaluate the following:
`int_0^pi x log sin x "d"x`
`int "dx"/(sin(x - "a")sin(x - "b"))` is equal to ______.
If `int(2e^(5x) + e^(4x) - 4e^(3x) + 4e^(2x) + 2e^x)/((e^(2x) + 4)(e^(2x) - 1)^2)dx = tan^-1(e^x/a) - 1/(b(e^(2x) - 1)) + C`, where C is constant of integration, then value of a + b is equal to ______.
`intsqrt(1+x) dx` = ______
`inte^(xloga).e^x dx` is ______
If u and v are two differentiable functions of x, then prove that `intu*v*dx = u*intv dx - int(d/dx u)(intv dx)dx`. Hence evaluate: `intx cos x dx`
