Advertisements
Advertisements
Question
Evaluate the following definite integrals: `int_2^3 x/(x^2 - 1)*dx`
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^(pi/2) cosx/((1 + sinx)(2 + sin x))*dx`
Evaluate the following : `int_((-pi)/4)^(pi/4) x^3 sin^4x*dx`
`int_0^(log5) (e^x sqrt(e^x - 1))/(e^x + 3) * dx` = ______.
Choose the correct option from the given alternatives :
The value of `int_((-pi)/4)^(pi/4) log((2+ sin theta)/(2 - sin theta))*d theta` is
Evaluate the following : `int_(-1)^(1) (1 + x^3)/(9 - x^2)*dx`
Evaluate the following definite integrals: `int_2^3 x/((x + 2)(x + 3)). dx`
Evaluate the following integrals : `int_0^1 log(1/x - 1)*dx`
Choose the correct alternative :
If `int_0^"a" 3x^2*dx` = 8, then a = ?
Choose the correct alternative :
`int_"a"^"b" f(x)*dx` =
Solve the following:
`int_0^1 e^(x^2)*x^3dx`
`int_1^2 ("e"^(1/x))/(x^2) "d"x` =
Evaluate `int_1^2 (3x)/((9x^2 - 1)) "d"x`
Evaluate `int_1^2 "e"^(2x) (1/x - 1/(2x^2)) "d"x`
Evaluate `int_0^"a" x^2 ("a" - x)^(3/2) "d"x`
`int_0^(pi/2) (cos x)/((4 + sin x)(3 + sin x))`dx = ?
`int_0^1 1/(2x + 5)dx` = ______
Evaluate the following definite integral:
`int_4^9 1/sqrt(x)dx`
`int_0^4 1/sqrt(4x - x^2)dx` = ______.
Solve the following.
`int_0^1 e^(x^2) x^3 dx`
Evaluate the following definite integral:
`int_1^2 (3x)/((9x^2 - 1))*dx`
