Advertisements
Advertisements
प्रश्न
Find `int_ (log "x")^2 d"x"`
Advertisements
उत्तर
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
संबंधित प्रश्न
Find the integrals of the function:
sin2 (2x + 5)
Find the integrals of the function:
sin 3x cos 4x
Find the integrals of the function:
cos 2x cos 4x cos 6x
Find the integrals of the function:
sin3 (2x + 1)
Find the integrals of the function:
sin x sin 2x sin 3x
Find the integrals of the function:
sin 4x sin 8x
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:
`(sin^3 x + cos^3 x)/(sin^2x cos^2 x)`
Find the integrals of the function:
`1/(sin xcos^3 x)`
`int (e^x(1 +x))/cos^2(e^x x) dx` equals ______.
Find `int (sin^2 x - cos^2x)/(sin x cos x) dx`
Evaluate `int_0^(3/2) |x sin pix|dx`
Find `int((3 sin x - 2) cos x)/(13 - cos^2 x- 7 sin x) dx`
Differentiate : \[\tan^{- 1} \left( \frac{1 + \cos x}{\sin x} \right)\] with respect to x .
Find `int_ (sin "x" - cos "x" )/sqrt(1 + sin 2"x") d"x", 0 < "x" < π / 2 `
Find `int_ (sin2"x")/((sin^2 "x"+1)(sin^2"x"+3))d"x"`
Find:
`int"dx"/sqrt(5-4"x" - 2"x"^2)`
Evaluate `int tan^8 x sec^4 x"d"x`
`int "dx"/(sin^2x cos^2x)` is equal to ______.
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`
Evaluate the following:
`int sin^-1 sqrt(x/("a" + x)) "d"x` (Hint: Put x = a tan2θ)
The value of the integral `int_(1/3)^1 (x - x^3)^(1/3)/x^4 dx` is
`int (cos^2x)/(sin x + cos x)^2 dx` is equal to
