Advertisements
Advertisements
Question
Evaluate `int x log x "d"x`
Advertisements
Solution
Let I = `int x* log x "d"x`
= `log x int x"d"x - int["d"/("d"x) (log x) int x"d"x] "d"x`
= `log x* x^2/2 - int[1/x xx x^2/2] "d"x`
= `x^2/2 log x - 1/2 int x "d"x`
= `x^2/2 log x - 1/2* x^2/2 + "c"`
∴ I = `x^2/2 log x - x^2/4 + "c"`
APPEARS IN
RELATED QUESTIONS
Integrate the rational function:
`x/((x-1)(x- 2)(x - 3))`
Integrate the rational function:
`x/((x^2+1)(x - 1))`
Find :
`∫ sin(x-a)/sin(x+a)dx`
Integrate the following w.r.t. x : `x^2/((x^2 + 1)(x^2 - 2)(x^2 + 3))`
Integrate the following w.r.t. x : `(12x + 3)/(6x^2 + 13x - 63)`
Integrate the following w.r.t. x : `(2x)/(4 - 3x - x^2)`
Integrate the following w.r.t. x : `2^x/(4^x - 3 * 2^x - 4`
Integrate the following w.r.t. x : `(3x - 2)/((x + 1)^2(x + 3)`
Integrate the following w.r.t. x : `(2log x + 3)/(x(3 log x + 2)[(logx)^2 + 1]`
Integrate the following w.r.t. x: `(2x^2 - 1)/(x^4 + 9x^2 + 20)`
Integrate the following with respect to the respective variable : `cot^-1 ((1 + sinx)/cosx)`
Integrate the following w.r.t.x : `sec^2x sqrt(7 + 2 tan x - tan^2 x)`
Evaluate:
`int x/((x - 1)^2(x + 2)) dx`
`int sqrt(4^x(4^x + 4)) "d"x`
`int 1/(x(x^3 - 1)) "d"x`
`int ((x^2 + 2))/(x^2 + 1) "a"^(x + tan^(-1_x)) "d"x`
`int (7 + 4x + 5x^2)/(2x + 3)^(3/2) dx`
`int sqrt((9 + x)/(9 - x)) "d"x`
`int sin(logx) "d"x`
`int sec^2x sqrt(tan^2x + tanx - 7) "d"x`
`int "e"^(sin^(-1_x))[(x + sqrt(1 - x^2))/sqrt(1 - x^2)] "d"x`
`int x^3tan^(-1)x "d"x`
`int ("d"x)/(x^3 - 1)`
Evaluate:
`int (5e^x)/((e^x + 1)(e^(2x) + 9)) dx`
Evaluate `int x^2"e"^(4x) "d"x`
`int x/((x - 1)^2 (x + 2)) "d"x`
If `int(sin2x)/(sin5x sin3x)dx = 1/3log|sin 3x| - 1/5log|f(x)| + c`, then f(x) = ______
Evaluate the following:
`int sqrt(tanx) "d"x` (Hint: Put tanx = t2)
The numerator of a fraction is 4 less than its denominator. If the numerator is decreased by 2 and the denominator is increased by 1, the denominator becomes eight times the numerator. Find the fraction.
Evaluate.
`int (5x^2 - 6x + 3) / (2x -3) dx`
