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:
sin3 x cos3 x
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:
`(1-cosx)/(1 + cos x)`
Find the integrals of the function:
sin4 x
Find the integrals of the function:
tan3 2x sec 2x
Find the integrals of the function:
sin−1 (cos x)
`int (sin^2x - cos^2 x)/(sin^2 x cos^2 x) dx` is equal to ______.
Find `int_ sin ("x" - a)/(sin ("x" + a )) d"x"`
Find the area of the triangle whose vertices are (-1, 1), (0, 5) and (3, 2), using integration.
Find: `int sec^2 x /sqrt(tan^2 x+4) dx.`
Find: `int sin^-1 (2x) dx.`
Evaluate `int tan^8 x sec^4 x"d"x`
Find `int x^2tan^-1x"d"x`
Evaluate the following:
`int tan^2x sec^4 x"d"x`
Evaluate the following:
`int (sinx + cosx)/sqrt(1 + sin 2x) "d"x`
Evaluate the following:
`int (sin^6x + cos^6x)/(sin^2x cos^2x) "d"x`
Evaluate the following:
`int (cosx - cos2x)/(1 - cosx) "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
What does integration using trigonometric identities mean?
Which identity is used for the product of sine and cosine?
Using \(\cos^2 x = \frac{1+\cos 2x}{2}\), which expression is equivalent to \(\int \cos^2 x\,dx\)?
What is \(\int \cos^2 x\,dx\)?
Applying the product-to-sum identity to \(\sin 2x\cos 3x\) gives which expression?
What must always be added to an indefinite integral?
