Advertisements
Advertisements
प्रश्न
Evaluate: `int sqrt("x"^2 + 2"x" + 5)` dx
Advertisements
उत्तर
Let I = `int sqrt("x"^2 + 2"x" + 5)` dx
`= int sqrt("x"^2 + 2"x" + 1 + 4)` dx
`= int sqrt(("x + 1")^2 + (2)^2)` dx
`= ("x" + 1)/2 sqrt(("x" + 1)^2 + (2)^2) + (2)^2/2 log |("x + 1") + sqrt(("x + 1")^2 + (2)^2)|` + c
∴ I = `("x" + 1)/2 sqrt("x"^2 + 2"x" + 5) + 2 log |("x + 1") + sqrt("x"^2 + 2"x" + 5)|` + c
APPEARS IN
संबंधित प्रश्न
Evaluate :`intxlogxdx`
Find `int((3sintheta-2)costheta)/(5-cos^2theta-4sin theta)d theta`.
Write a value of
Write a value of\[\int\sqrt{x^2 - 9} \text{ dx}\]
Evaluate: \[\int\frac{x^3 - 1}{x^2} \text{ dx}\]
Integrate the following functions w.r.t. x : `e^(3x)/(e^(3x) + 1)`
Integrate the following functions w.r.t. x : `(1)/(x(x^3 - 1)`
Integrate the following functions w.r.t. x : `(cos3x - cos4x)/(sin3x + sin4x)`
Integrate the following functions w.r.t. x : `(sinx + 2cosx)/(3sinx + 4cosx)`
Choose the correct options from the given alternatives :
`int dx/(cosxsqrt(sin^2x - cos^2x))*dx` =
`int logx/(log ex)^2*dx` = ______.
Evaluate the following.
`int 1/("x" log "x")`dx
Evaluate the following.
`int 1/(x(x^6 + 1))` dx
Evaluate the following.
`int 1/(sqrt("x"^2 + 4"x"+ 29))` dx
Choose the correct alternative from the following.
The value of `int "dx"/sqrt"1 - x"` is
Choose the correct alternative from the following.
`int "dx"/(("x" - "x"^2))`=
Evaluate `int "x - 1"/sqrt("x + 4")` dx
`int x/(x + 2) "d"x`
If `tan^-1x = 2tan^-1((1 - x)/(1 + x))`, then the value of x is ______
`int (cos x)/(1 - sin x) "dx" =` ______.
`int ("d"x)/(sinx cosx + 2cos^2x)` = ______.
`int x/sqrt(1 - 2x^4) dx` = ______.
(where c is a constant of integration)
Evaluate `int (1+x+x^2/(2!))dx`
Evaluate `int(1+ x + x^2/(2!)) dx`
Evaluate the following.
`int "x"^3/sqrt(1 + "x"^4)` dx
Evaluate the following.
`int1/(x^2 + 4x - 5) dx`
Evaluate `int(1+x+x^2/(2!))dx`
If f'(x) = 4x3 - 3x2 + 2x + k, f(0) = 1 and f(1) = 4, find f(x).
