Advertisements
Advertisements
Question
Evaluate: `int_1^2 x/(1 + x^2) "d"x`
Advertisements
Solution
`int_1^2 x/(1 + x^2) "d"x = 1/2 int_1^2 (2x)/(1 + x^2) "d"x`
= `1/2[log|1 + x^2|]_1^2` ........`[∵ int ("f'"(x))/("f"(x)) "d"x = log|"f"(x)| + c"]`
= `1/2(log 5 - log 2)`
=`1/2 log(5/2)`
APPEARS IN
RELATED QUESTIONS
Evaluate: `int_0^(pi/2) x sin x.dx`
Evaluate: `int_0^oo xe^-x.dx`
Evaluate: `int_0^π sin^3x (1 + 2cosx)(1 + cosx)^2.dx`
`int_0^(x/4) sqrt(1 + sin 2x) "d"x` =
`int_(pi/5)^((3pi)/10) sinx/(sinx + cosx) "d"x` =
Let I1 = `int_"e"^("e"^2) 1/logx "d"x` and I2 = `int_1^2 ("e"^x)/x "d"x` then
Evaluate: `int_0^1 1/(1 + x^2) "d"x`
Evaluate: `int_0^(pi/2) (sin2x)/(1 + sin^2x) "d"x`
Evaluate: `int_(pi/6)^(pi/3) sin^2 x "d"x`
Evaluate:
`int_0^(pi/2) cos^3x dx`
Evaluate: `int_0^pi cos^2 x "d"x`
Evaluate: `int_0^(pi/4) (tan^3x)/(1 + cos 2x) "d"x`
Evaluate: `int_0^(pi/4) cosx/(4 - sin^2 x) "d"x`
Evaluate: `int_0^9 sqrt(x)/(sqrt(x) + sqrt(9 - x) "d"x`
Evaluate: `int_0^(pi/2) (sin^4x)/(sin^4x + cos^4x) "d"x`
Evaluate: `int_(-1)^1 |5x - 3| "d"x`
Evaluate: `int_0^(pi/4) sec^4x "d"x`
Evaluate: `int_0^(pi/2) 1/(5 + 4cos x) "d"x`
Evaluate: `int_0^(pi/2) cos x/((1 + sinx)(2 + sinx)) "d"x`
Evaluate: `int_(-1)^1 1/("a"^2"e"^x + "b"^2"e"^(-x)) "d"x`
Evaluate: `int_0^3 x^2 (3 - x)^(5/2) "d"x`
Evaluate: `int_0^(1/2) 1/((1 - 2x^2) sqrt(1 - x^2)) "d"x`
Evaluate: `int_(1/sqrt(2))^1 (("e"^(cos^-1x))(sin^-1x))/sqrt(1 - x^2) "d"x`
Evaluate: `int_0^pi x*sinx*cos^2x* "d"x`
Evaluate: `int_0^(pi/4) (cos2x)/(1 + cos 2x + sin 2x) "d"x`
Evaluate: `int_0^(pi/4) log(1 + tanx) "d"x`
Evaluate: `int_0^pi 1/(3 + 2sinx + cosx) "d"x`
Evaluate: `int_0^(π/4) sec^4 x dx`
Evaluate:
`int_(π/4)^(π/2) cot^2x dx`.
Evaluate: `int_0^1 tan^-1(x/sqrt(1 - x^2))dx`.
Evaluate:
`int_0^(π/2) sin^8x dx`
Evaluate:
`int_(-π/2)^(π/2) |sinx|dx`
Evaluate:
`int_-4^5 |x + 3|dx`
Evaluate:
`int_0^(π/2) (sin 2x)/(1 + sin^4x)dx`
`int_0^1 x^2/(1 + x^2)dx` = ______.
Evaluate:
`int_0^(π/2) sinx/(1 + cosx)^3 dx`
If `int_0^π f(sinx)dx = kint_0^π f(sinx)dx`, then find the value of k.
