Advertisements
Advertisements
प्रश्न
Fill in the Blank.
`int 1/"x"^3 [log "x"^"x"]^2 "dx" = "P" (log "x")^3` + c, then P = _______
उत्तर
`int 1/"x"^3 [log "x"^"x"]^2 "dx" = "P" (log "x")^3` + c, then P = `underline(1/3)`
Explanation:
Let I = `int 1/"x"^3 [log "x"^"x"]^2 "dx" = "P" * (log "x")^3` + c
I = `int 1/"x"^3 [log "x"^"x"]^2 "dx" = int 1/"x"^3 ("x log x")^2 * "dx"`
`=int 1/"x"^3 * "x"^2 * (log "x")^2 "dx" = int 1/"x" (log "x")^2 * "dx"`
∴ Put log x = t
∴ `1/"x"` dx = dt
∴ I = `int "t"^2 * "dt"`
`= "t"^3/3 + "c"`
`= 1/3 (log "x")^3 + "c"`
∴ P = `1/3`
APPEARS IN
संबंधित प्रश्न
Integrate the functions:
`1/(1 + cot x)`
`(10x^9 + 10^x log_e 10)/(x^10 + 10^x) dx` equals:
Write a value of\[\int\frac{\sin 2x}{a^2 \sin^2 x + b^2 \cos^2 x} \text{ dx }\]
Write a value of\[\int e^x \left( \frac{1}{x} - \frac{1}{x^2} \right) dx\] .
Show that : `int _0^(pi/4) "log" (1+"tan""x")"dx" = pi /8 "log"2`
Evaluate the following integrals : `int (cos2x)/(sin^2x.cos^2x)dx`
Evaluate the following integrals : `int tanx/(sec x + tan x)dx`
Integrate the following functions w.r.t. x : `(4e^x - 25)/(2e^x - 5)`
Integrate the following functions w.r.t. x : tan5x
Evaluate the following integrals : `int (3x + 4)/sqrt(2x^2 + 2x + 1).dx`
Evaluate the following.
`int 1/("x"^2 + 4"x" - 5)` dx
Evaluate: If f '(x) = `sqrt"x"` and f(1) = 2, then find the value of f(x).
`int 1/(xsin^2(logx)) "d"x`
`int x^3 e^(x^2) dx`
The value of `int ("d"x)/(sqrt(1 - x))` is ______.
Evaluate `int(1+x+x^2/(2!))dx`
Evaluate `int1/(x(x-1))dx`
Evaluate the following.
`int1/(x^2 + 4x - 5)dx`
