Advertisements
Advertisements
प्रश्न
Integrate the functions:
(4x + 2) `sqrt(x^2 + x +1)`
उत्तर
Let `I = int (4x + 2) sqrt(x^2 + x + 1)` dx
or `I = 2 int (2x + 1) sqrt ((x^2 + x + 1))` dx
Taking x2 + x + 1 = t
2x + 1 = dt
Hence, `I = 2 int sqrt t dt`
`= 2 int t^(1/2) dt = 2. 2/3 t^(3/2) + C`
`= 4/3 (x^2 + x + 1)^(3/2) + C`
APPEARS IN
संबंधित प्रश्न
Evaluate : `int(x-3)sqrt(x^2+3x-18) dx`
Find `int((3sintheta-2)costheta)/(5-cos^2theta-4sin theta)d theta`
Integrate the functions:
`(2x)/(1 + x^2)`
Integrate the functions:
`cos sqrt(x)/sqrtx`
Evaluate : `∫1/(3+2sinx+cosx)dx`
Write a value of
Write a value of
Write a value of\[\int\frac{1}{x \left( \log x \right)^n} \text { dx }\].
Integrate the following functions w.r.t. x : `(sin6x)/(sin 10x sin 4x)`
Integrate the following functions w.r.t. x : `(sinx cos^3x)/(1 + cos^2x)`
Evaluate the following integrals : `int (3x + 4)/sqrt(2x^2 + 2x + 1).dx`
Evaluate the following integrals : `int sqrt((x - 7)/(x - 9)).dx`
Evaluate the following integrals : `int (3cosx)/(4sin^2x + 4sinx - 1).dx`
Choose the correct option from the given alternatives :
`int (1 + x + sqrt(x + x^2))/(sqrt(x) + sqrt(1 + x))*dx` =
Evaluate the following.
`int ("2x" + 6)/(sqrt("x"^2 + 6"x" + 3))` dx
Choose the correct alternative from the following.
`int "x"^2 (3)^("x"^3) "dx"` =
Fill in the Blank.
To find the value of `int ((1 + log "x") "dx")/"x"` the proper substitution is ________
State whether the following statement is True or False.
If ∫ x f(x) dx = `("f"("x"))/2`, then find f(x) = `"e"^("x"^2)`
Evaluate: `int "x" * "e"^"2x"` dx
`int (2(cos^2 x - sin^2 x))/(cos^2 x + sin^2 x)` dx = ______________
`int logx/x "d"x`
`int x/(x + 2) "d"x`
To find the value of `int ((1 + logx))/x` dx the proper substitution is ______
State whether the following statement is True or False:
`int"e"^(4x - 7) "d"x = ("e"^(4x - 7))/(-7) + "c"`
`int (cos x)/(1 - sin x) "dx" =` ______.
`int[ tan (log x) + sec^2 (log x)] dx= ` ______
The integral `int ((1 - 1/sqrt(3))(cosx - sinx))/((1 + 2/sqrt(3) sin2x))dx` is equal to ______.
`int(1 - x)^(-2)` dx = `(1 - x)^(-1) + c`
Evaluate.
`int(5"x"^2 - 6"x" + 3)/(2"x" - 3) "dx"`
Evaluate the following
`int x^3/sqrt(1+x^4) dx`
Evaluate:
`int sin^2(x/2)dx`
Evaluate the following.
`intxsqrt(1+x^2)dx`
Evaluate `int1/(x(x-1))dx`
Evaluate:
`int(5x^2-6x+3)/(2x-3)dx`
Evaluate `int1/(x(x-1))dx`
Evaluate the following.
`int1/(x^2 + 4x - 5) dx`
Evaluate:
`intsqrt(sec x/2 - 1)dx`