Advertisements
Advertisements
Question
Find `int dx/(x^2 + 4x + 8)`
Solution
Consider the integral, `int dx/(x^2 + 4x + 8)`
I = `int 1/(x^2 + 4x + 8)` dx
= `int 1/(x^2 + 4x + 4 + 4)` dx
= `int 1/((x + 2)^2 + 2^2) dx` {Use Intergral Formula : `int 1/(x^2 + a^2) dx = 1/a tan^(-1) (x/a)}`
`= 1/2 tan^(-1) (x + 2)/2 + C`
APPEARS IN
RELATED QUESTIONS
Evaluate :`int_(pi/6)^(pi/3) dx/(1+sqrtcotx)`
Find the integrals of the function:
sin3 x cos3 x
Find the integrals of the function:
`(1-cosx)/(1 + cos x)`
Find the integrals of the function:
`(sin^2 x)/(1 + cos x)`
Find the integrals of the function:
`(cos 2x - cos 2 alpha)/(cos x - cos alpha)`
Find the integrals of the function:
`(cos x - sinx)/(1+sin 2x)`
Find the integrals of the function:
`1/(sin xcos^3 x)`
Find the integrals of the function:
`(cos 2x)/(cos x + sin x)^2`
Find the integrals of the function:
sin−1 (cos x)
Find the integrals of the function:
`1/(cos(x - a) cos(x - b))`
Find `int (2x)/((x^2 + 1)(x^4 + 4))`dx
Differentiate : \[\tan^{- 1} \left( \frac{1 + \cos x}{\sin x} \right)\] with respect to x .
Evaluate : \[\int\limits_0^\pi \frac{x \tan x}{\sec x \cdot cosec x}dx\] .
Find `int_ (sin "x" - cos "x" )/sqrt(1 + sin 2"x") d"x", 0 < "x" < π / 2 `
Find:
`int"dx"/sqrt(5-4"x" - 2"x"^2)`
Integrate the function `cos("x + a")/sin("x + b")` w.r.t. x.
Evaluate `int tan^8 x sec^4 x"d"x`
`int "dx"/(sin^2x cos^2x)` is equal to ______.
`int (sin^6x)/(cos^8x) "d"x` = ______.
Evaluate the following:
`int ((1 + cosx))/(x + sinx) "d"x`
Evaluate the following:
`int (sinx + cosx)/sqrt(1 + sin 2x) "d"x`
Evaluate the following:
`int sqrt(1 + sinx)"d"x`
Evaluate the following:
`int (cosx - cos2x)/(1 - cosx) "d"x`
`int (cos^2x)/(sin x + cos x)^2 dx` is equal to
