Advertisements
Advertisements
प्रश्न
To find the value of `int ((1 + log x) )/x dx` the proper substitution is ______.
Advertisements
उत्तर
To find the value of `int ((1 + log x))/x dx` the proper substitution is 1 + log x = t.
Explanation:
Given integral is `int ((1 + log x) )/x dx`
Let t = 1 + log x
Differentiate w.r.t. x we get,
`dt/dx = 1/x`
⇒ dx = x dt
Now substitute into the integral:
`int t/x dx`
= `int t/x * x dt`
= `∫ t dt`
संबंधित प्रश्न
Integrate the functions:
`xsqrt(1+ 2x^2)`
Write a value of\[\int\frac{\sin x + \cos x}{\sqrt{1 + \sin 2x}} dx\]
The value of \[\int\frac{1}{x + x \log x} dx\] is
Integrate the following functions w.r.t. x : `((sin^-1 x)^(3/2))/(sqrt(1 - x^2)`
Integrate the following functions w.r.t. x : `(x^2 + 2)/((x^2 + 1)).a^(x + tan^-1x)`
Integrate the following functions w.r.t.x:
`(5 - 3x)(2 - 3x)^(-1/2)`
Integrate the following functions w.r.t. x : `(cos3x - cos4x)/(sin3x + sin4x)`
Integrate the following functions w.r.t. x : `(sin6x)/(sin 10x sin 4x)`
Integrate the following functions w.r.t. x : `int (1)/(3 + 2sin x - cosx)dx`
Integrate the following functions w.r.t. x : `int (1)/(cosx - sinx).dx`
Evaluate the following integral:
`int (3cosx)/(4sin^2x + 4sinx - 1).dx`
If f'(x) = x2 + 5 and f(0) = −1, then find the value of f(x).
Evaluate the following.
`int 1/("a"^2 - "b"^2 "x"^2)` dx
Choose the correct alternative from the following.
The value of `int "dx"/sqrt"1 - x"` is
Evaluate: `int "x" * "e"^"2x"` dx
Evaluate: `int sqrt("x"^2 + 2"x" + 5)` dx
`int "e"^x[((x + 3))/((x + 4)^2)] "d"x`
`int 1/(xsin^2(logx)) "d"x`
`int(1 - x)^(-2) dx` = ______.
`int sin^-1 x`dx = ?
`int sec^6 x tan x "d"x` = ______.
`int ("d"x)/(x(x^4 + 1))` = ______.
Find : `int sqrt(x/(1 - x^3))dx; x ∈ (0, 1)`.
Evaluate `int (1)/(x(x - 1))dx`
Evaluate.
`int (5x^2 - 6x + 3)/(2x - 3) dx`
Evaluate `int 1/(x(x-1))dx`
If f'(x) = 4x3- 3x2 + 2x + k, f(0) = 1 and f(1) = 4, find f(x).
Evaluate the following.
`intx^3/sqrt(1+x^4)dx`
Which substitution is appropriate for \[\sqrt{\frac{x}{x-a}}\], \[\sqrt{\frac{x-a}{x}}\], \[\sqrt{x(x-\mathrm{a})}\], or \[\frac{1}{\sqrt{x(x-\mathrm{a})}}\]?
For \[\sqrt{\frac{x-\alpha}{\beta-x}}\] or \[\sqrt{(x-\alpha)(\beta-x)}\], where \[\beta>\alpha\], which substitution is used?
