Advertisements
Advertisements
Question
Choose the correct alternative :
`int_2^3 x/(x^2 - 1)*dx` =
Options
`log (8/3)`
`-log (8/3)`
`(1)/(2)log(8/3)`
`(-1)/(2)log(8/3)`
Solution
Let I = `int_2^3 x/(x^2 - 1)*dx`
Put x2 – 1 = t
∴ 2x·dx = dt
∴ x·dx = `(1)/(2)*dt`
When x = 2, t = 22 – 1 = 3
When x = 3, t = 32 – 1 = 8
∴ I = `int_3^8 (1)/"t"*"dt"/(2)`
= `(1)/(2)int_3^8 "dt"/"t"`
= `(1)/(2)[log |"t"|]_3^8`
= `(1)/(2)(log 8 - log 3)`
∴ I = `(1)/(2) log (8/3)`.
APPEARS IN
RELATED QUESTIONS
Evaluate : `int_0^4 (1)/sqrt(4x - x^2)*dx`
Evaluate : `int_0^(pi/2) (sinx - cosx)/(1 + sinx cosx)*dx`
Evaluate the following : `int_0^(pi/4) (tan^3x)/(1 +cos2x)*dx`
Evaluate the following : `int_0^pi (sin^-1x + cos^-1x)^3 sin^3x*dx`
Evaluate the following : if `int_a^a sqrt(x)*dx = 2a int_0^(pi/2) sin^3x*dx`, find the value of `int_a^(a + 1)x*dx`
Fill in the blank : `int_4^9 (1)/sqrt(x)*dx` = _______
State whether the following is True or False : `int_"a"^"b" f(x)*dx = int_(-"b")^(-"a") f(x)*dx`
State whether the following is True or False : `int_1^2 sqrt(x)/(sqrt(3 - x) + sqrt(x))*dx = (1)/(2)`
Solve the following : `int_2^3 x/((x + 2)(x + 3))*dx`
Solve the following:
`int_1^3 x^2 log x*dx`
Solve the following : `int_(-2)^3 (1)/(x + 5)*dx`
Solve the following : `int_0^1 (x^2 + 3x + 2)/sqrt(x)*dx`
Solve the following : `int_0^9 (1)/(1 + sqrt(x))*dx`
`int_1^2 ("e"^(1/x))/(x^2) "d"x` =
`int_0^"a" 4x^3 "d"x` = 81, then a = ______
If `int_0^"a" (2x + 1) "d"x` = 2, find a
`int_((-pi)/8)^(pi/8) log ((2 - sin x)/(2 + sin x))` dx = ______.
`int_0^1 tan^-1 ((2x - 1)/(1 + x - x^2))` dx = ?
Evaluate the following definite integral:
`int_1^2 (3x)/((9x^2 - 1))dx`
Evaluate the following definite intergral:
`int_1^3 log x·dx`