Advertisements
Advertisements
Question
Solve the following : `int_0^4 (1)/sqrt(x^2 + 2x + 3)*dx`
Solution
Let I = `int_0^4 (1)/sqrt(x^2 + 2x + 3)*dx`
= `int_0^4 (1)/sqrt(x^2 + 2x + 1 - 1 + 3)*dx`
= `int_0^4 (1)/((sqrt(x + 1))^2 + 2)*dx`
= `int_0^4 (1)/sqrt((x + 1)^2 + (sqrt(2))^2)*dx`
= `[log |x + 1 + sqrt((x + 1)62 + (sqrt(2))^2|]_0^4`
= `log |5 + sqrt(27)| - log| 1 + sqrt(3)|`
= `log |5 + 3sqrt(3)| - log| 1 + sqrt(3)|`
∴ I = `log |(5 + 3sqrt(3))/(1 + sqrt(3))|`.
APPEARS IN
RELATED QUESTIONS
Evaluate : `int_0^(pi/4) sin^4x*dx`
Evaluate : `int_0^(1/sqrt(2)) (sin^-1x)/(1 - x^2)^(3/2)*dx`
Evaluate : `int _((1)/(sqrt(2)))^1 (e^(cos^-1x) sin^-1x)/(sqrt(1 - x^2))*dx`
Evaluate the following : `int_(-a)^(a) (x + x^3)/(16 - x^2)*dx`
Evaluate the following : `int_0^(pi/2) cosx/(3cosx + sinx)*dx`
Evaluate the following : `int_(-1)^(1) (1 + x^3)/(9 - x^2)*dx`
Evaluate the following : `int_0^1 sin^-1 ((2x)/(1 + x^2))*dx`
Evaluate the following definite integrals: `int_0^1 (1)/(sqrt(1 + x) + sqrt(x))*dx`
Fill in the blank : If `int_0^"a" 3x^2*dx` = 8, then a = _______
State whether the following is True or False : `int_"a"^"b" f(x)*dx = int_(-"b")^(-"a") f(x)*dx`
Solve the following : `int_0^9 (1)/(1 + sqrt(x))*dx`
Choose the correct alternative:
`int_2^3 x/(x^2 - 1) "d"x` =
Evaluate `int_1^3 log x "d"x`
Evaluate:
`int_(-π/2)^(π/2) (sin^3x)/(1 + cos^2x)dx`
Evaluate the following definite intergral:
`int_4^9 1/sqrt(x)dx`
Evaluate the following definite intergral:
`int_1^2 (3x)/((9x^2-1 )`dx
Evaluate the following integral:
`int_0^1 x(1-x)^5 dx`
Evaluate the following definite integral:
`int_1^2 (3x)/((9x^2 - 1)) dx`
Evaluate the following definite integral:
`int_1^2 (3x)/((9x^2 - 1)) dx`
Evaluate the following definite intergral:
`int_-2^3 1/(x+5).dx`
