Advertisements
Advertisements
प्रश्न
Find `int dx/(x^2 + 4x + 8)`
Advertisements
उत्तर
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
संबंधित प्रश्न
Evaluate :`int_(pi/6)^(pi/3) dx/(1+sqrtcotx)`
Evaluate : `intsin(x-a)/sin(x+a)dx`
Find the integrals of the function:
`(1-cosx)/(1 + cos x)`
Find the integrals of the function:
cos4 2x
Find the integrals of the function:
`(cos x - sinx)/(1+sin 2x)`
Find the integrals of the function:
tan3 2x sec 2x
Find the integrals of the function:
`(cos 2x)/(cos x + sin x)^2`
Find `int (sin^2 x - cos^2x)/(sin x cos x) dx`
Evaluate: `int_0^π (x sin x)/(1 + cos^2x) dx`.
Evaluate `int_0^(3/2) |x sin pix|dx`
Find `int((3 sin x - 2) cos x)/(13 - cos^2 x- 7 sin x) dx`
Differentiate : \[\tan^{- 1} \left( \frac{1 + \cos x}{\sin x} \right)\] with respect to x .
Find `int_ (sin "x" - cos "x" )/sqrt(1 + sin 2"x") d"x", 0 < "x" < π / 2 `
Find `int_ (sin2"x")/((sin^2 "x"+1)(sin^2"x"+3))d"x"`
Find: `int_ (cos"x")/((1 + sin "x") (2+ sin"x")) "dx"`
Find:
`int"dx"/sqrt(5-4"x" - 2"x"^2)`
Find: `intsqrt(1 - sin 2x) dx, pi/4 < x < pi/2`
Find: `int sin^-1 (2x) dx.`
`int (sin^6x)/(cos^8x) "d"x` = ______.
Evaluate the following:
`int ("d"x)/(1 + cos 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 (sin^6x + cos^6x)/(sin^2x cos^2x) "d"x`
Evaluate the following:
`int (cosx - cos2x)/(1 - cosx) "d"x`
`int (cos^2x)/(sin x + cos x)^2 dx` is equal to
