Advertisements
Advertisements
Question
Find: `int logx/(1 + log x)^2 dx`
Advertisements
Solution
`int logx/(1 + log x)^2 dx = int (log x + 1 - 1)/(1 + log x)^2 dx`
= `int 1/(1 + log x) dx - int 1/(1 + log x)^2 dx`
= `1/(1 + log x) xx x - int (-1)/(1 + log x)^2 xx 1/x xx xdx - int 1/(1 + log x)^2 dx`
= ` x/(1 + log x) + c`
APPEARS IN
RELATED QUESTIONS
Evaluate the following integral:
Evaluate each of the following integral:
Solve each of the following integral:
Write the coefficient a, b, c of which the value of the integral
If \[\int\limits_0^1 f\left( x \right) dx = 1, \int\limits_0^1 xf\left( x \right) dx = a, \int\limits_0^1 x^2 f\left( x \right) dx = a^2 , then \int\limits_0^1 \left( a - x \right)^2 f\left( x \right) dx\] equals
\[\int\limits_{- \pi/4}^{\pi/4} \left| \tan x \right| dx\]
\[\int\limits_0^\pi \frac{x \tan x}{\sec x + \tan x} dx\]
Evaluate the following:
`int_1^4` f(x) dx where f(x) = `{{:(4x + 3",", 1 ≤ x ≤ 2),(3x + 5",", 2 < x ≤ 4):}`
Evaluate the following:
`Γ (9/2)`
Choose the correct alternative:
`int_0^1 (2x + 1) "d"x` is
Choose the correct alternative:
`Γ(3/2)`
Evaluate `int sqrt((1 + x)/(1 - x)) "d"x`, x ≠1
Evaluate `int (x^2"d"x)/(x^4 + x^2 - 2)`
If x = `int_0^y "dt"/sqrt(1 + 9"t"^2)` and `("d"^2y)/("d"x^2)` = ay, then a equal to ______.
