Advertisements
Advertisements
प्रश्न
Evaluate the following:
`int ("e"^(6logx) - "e"^(5logx))/("e"^(4logx) - "e"^(3logx)) "d"x`
Advertisements
उत्तर
Let I = `int ("e"^(6logx) - "e"^(5logx))/("e"^(4logx) - "e"^(3logx)) "d"x`
= `int ("e"^(log x^6) - "e"^(log x^5))/("e"^(logx^4) - "e"^(log x^3)) "d"x` .....[∵ a log b – log ba]
= `int (x^6 - x^5)/(x^4 - x^3) "d"x` .....[∵ elogx = x]
= `int (x^3 - x^2)/(x - 1) "d"x`
= `int (x^2(x - 1))/(x - 1) "d"x`
= `int x^2 "d"x`
= `x^3/3 + "C"`
APPEARS IN
संबंधित प्रश्न
Evaluate :`int_0^(pi/2)1/(1+cosx)dx`
Evaluate : `int1/(3+5cosx)dx`
Evaluate `∫_0^(3/2)|x cosπx|dx`
Evaluate :
`∫_(-pi)^pi (cos ax−sin bx)^2 dx`
find `∫_2^4 x/(x^2 + 1)dx`
Evaluate the integral by using substitution.
`int_0^1 x/(x^2 +1)`dx
Evaluate the integral by using substitution.
`int_0^2 xsqrt(x+2)` (Put x + 2 = `t^2`)
Evaluate of the following integral:
Evaluate of the following integral:
Evaluate :
Evaluate:
Evaluate:
Evaluate:
Evaluate the following integral:
Evaluate the following integral:
Evaluate the following integral:
Evaluate the following integral:
Evaluate the following integral:
Evaluate each of the following integral:
Evaluate the following integral:
Evaluate the following integral:
Evaluate
\[\int\limits_0^\pi \frac{x}{1 + \sin \alpha \sin x}dx\]
Evaluate the following integral:
Evaluate : \[\int\limits_{- 2}^1 \left| x^3 - x \right|dx\] .
Find : \[\int\frac{x \sin^{- 1} x}{\sqrt{1 - x^2}}dx\] .
Evaluate: `int_-π^π (1 - "x"^2) sin "x" cos^2 "x" d"x"`.
`int_(pi/5)^((3pi)/10) [(tan x)/(tan x + cot x)]`dx = ?
If `I_n = int_0^(pi/4) tan^n theta "d"theta " then " I_8 + I_6` equals ______.
`int_0^(pi4) sec^4x "d"x` = ______.
The value of `int_0^1 (x^4(1 - x)^4)/(1 + x^2) dx` is
Evaluate:
`int (1 + cosx)/(sin^2x)dx`
If `int x^5 cos (x^6)dx = k sin (x^6) + C`, find the value of k.
