Advertisements
Advertisements
Question
Find `int_ (log "x")^2 d"x"`
Advertisements
Solution
Let `I = int (log "x")^2 d"x"`
⇒ `I = int_ 1·(log "x")^2 d"x"`
⇒ `I = "x"·(log "x")^2 - int_ (2"x" log"x")/"x" d"x"`
⇒ `I = "x"·(log "x")^2 - I_1 + c_1` .....(i)
`I_1 = int_ 2·log "x"d"x"`
⇒ `I_1 = 2"x"· log"x"- 2 int_ "x"/"x" d"x"`
⇒ `I_1 = 2"x"·log "x" - 2"x" + c_2` .....(ii)
From (i) and (ii), we get
`I = "x"·(log "x")^2 - 2"x"·log "x"+ 2"x" + c_1 - c_2`
`I = "x"·(log "x")^2 - 2"x"·log "x"+ 2"x" + C` ...(where C = C1 - C2)
APPEARS IN
RELATED QUESTIONS
Find the integrals of the function:
cos 2x cos 4x cos 6x
Find the integrals of the function:
`(1-cosx)/(1 + cos x)`
Find the integrals of the function:
`cos x/(1 + cos x)`
Find the integrals of the function:
sin4 x
Find the integrals of the function:
`(cos 2x - cos 2 alpha)/(cos x - cos alpha)`
Find the integrals of the function:
tan3 2x sec 2x
Find the integrals of the function:
tan4x
Find the integrals of the function:
`(cos 2x+ 2sin^2x)/(cos^2 x)`
Find the integrals of the function:
sin−1 (cos x)
Find `int dx/(x^2 + 4x + 8)`
Find `int (2x)/((x^2 + 1)(x^4 + 4))`dx
Evaluate : \[\int\limits_0^\pi \frac{x \tan x}{\sec x \cdot cosec x}dx\] .
Find the area of the triangle whose vertices are (-1, 1), (0, 5) and (3, 2), using integration.
Find: `int_ (cos"x")/((1 + sin "x") (2+ sin"x")) "dx"`
Integrate the function `cos("x + a")/sin("x + b")` w.r.t. x.
Evaluate `int tan^8 x sec^4 x"d"x`
Find `int x^2tan^-1x"d"x`
`int "e"^x (cosx - sinx)"d"x` is equal to ______.
`int "dx"/(sin^2x cos^2x)` is equal to ______.
`int (sin^6x)/(cos^8x) "d"x` = ______.
Evaluate the following:
`int ((1 + cosx))/(x + sinx) "d"x`
Evaluate the following:
`int ("d"x)/(1 + cos x)`
Evaluate the following:
`int (sinx + cosx)/sqrt(1 + sin 2x) "d"x`
Evaluate the following:
`int (cosx - cos2x)/(1 - cosx) "d"x`
Evaluate the following:
`int "e"^(tan^-1x) ((1 + x + x^2)/(1 + x^2)) "d"x`
