Advertisements
Advertisements
Question
Integrate the following functions w.r.t. x : `(x + 1) sqrt(2x^2 + 3)`
Advertisements
Solution
Let I = `int (x + 1)sqrt(2x^2 + 3)`
Let x + 1 = `"A"[d/dx (2x^2 + 3)] + "B"`
= A (4x) + B
= 4Ax + B
Comparing the coefficients of and constant on both sides, we get
4A = 1, B = 1
∴ A = `(1)/(4), "B"` = 1
∴ x + 1 = `(1)/(4)(4x) + 1`
∴ I = `int [1/4 (4x) + 1]sqrt(2x^2 + 3).dx`
= `(1)/(4) int 4x sqrt(2x^2 + 3).dx + int sqrt(2x^2 + 3).dx`.
= I1 + I2
In I1 = put 2x2 + 3 = t
∴ 4x.dx = dt
∴ I1 = `(1)/(4) int t^(12).dt`
= `(1)/(4)(t^(3/2)/(3/2)) + c_1`
= `(1)/(6)(2x^2 + 3)^(3/2) + c_1`
I2 = `int sqrt(2x^2 + 3).dx`
= `sqrt(2) int sqrt(x^2 + 3/2).dx`
= `sqrt(2)[x/2sqrt(x^2 + 3/2) + ((3/2))/(2)log|x + sqrt(x^2 + 3/2)|] + c_2`
= `sqrt(2)[x/2sqrt(x^2 + 3/2) + (3)/(4)log|x + sqrt(x^2 + 3/2)|] + c_2`
∴ I = `(1)/(6)(2x^2 + 3)^(3/2) + sqrt(2)[x/2 sqrt(x^2 + 3/2) + (3)/(4) log|x + sqrt(x^2 + 3/2)|] + c`, where c = c1 + c2.
APPEARS IN
RELATED QUESTIONS
Integrate the function in `(xe^x)/(1+x)^2`.
Integrate the function in `e^x (1/x - 1/x^2)`.
`intx^2 e^(x^3) dx` equals:
Evaluate the following : `int x^2tan^-1x.dx`
Evaluate the following : `int x^3.tan^-1x.dx`
Evaluate the following: `int x.sin^-1 x.dx`
Evaluate the following : `int(sin(logx)^2)/x.log.x.dx`
Integrate the following functions w.r.t. x:
sin (log x)
Integrate the following functions w.r.t. x : `xsqrt(5 - 4x - x^2)`
Integrate the following functions w.r.t. x : `[x/(x + 1)^2].e^x`
Choose the correct options from the given alternatives :
`int (1)/(x + x^5)*dx` = f(x) + c, then `int x^4/(x + x^5)*dx` =
Integrate the following with respect to the respective variable : cos 3x cos 2x cos x
Evaluate the following.
`int e^x (1/x - 1/x^2)`dx
`int ("x" + 1/"x")^3 "dx"` = ______
Choose the correct alternative from the following.
`int (1 - "x")^(-2) "dx"` =
Evaluate: Find the primitive of `1/(1 + "e"^"x")`
Evaluate: `int ("ae"^("x") + "be"^(-"x"))/("ae"^("x") - "be"^(−"x"))` dx
Evaluate: `int "dx"/sqrt(4"x"^2 - 5)`
Evaluate: `int "e"^"x"/(4"e"^"2x" -1)` dx
`int 1/(4x + 5x^(-11)) "d"x`
`int sin4x cos3x "d"x`
`int(x + 1/x)^3 dx` = ______.
Evaluate `int 1/(4x^2 - 1) "d"x`
Evaluate `int (2x + 1)/((x + 1)(x - 2)) "d"x`
∫ log x · (log x + 2) dx = ?
`int_0^"a" sqrt("x"/("a" - "x")) "dx"` = ____________.
`int cot "x".log [log (sin "x")] "dx"` = ____________.
`int log x * [log ("e"x)]^-2` dx = ?
`int 1/sqrt(x^2 - 9) dx` = ______.
`int(logx)^2dx` equals ______.
If `int(2e^(5x) + e^(4x) - 4e^(3x) + 4e^(2x) + 2e^x)/((e^(2x) + 4)(e^(2x) - 1)^2)dx = tan^-1(e^x/a) - 1/(b(e^(2x) - 1)) + C`, where C is constant of integration, then value of a + b is equal to ______.
If `π/2` < x < π, then `intxsqrt((1 + cos2x)/2)dx` = ______.
Find `int (sin^-1x)/(1 - x^2)^(3//2) dx`.
Solve the differential equation (x2 + y2) dx - 2xy dy = 0 by completing the following activity.
Solution: (x2 + y2) dx - 2xy dy = 0
∴ `dy/dx=(x^2+y^2)/(2xy)` ...(1)
Puty = vx
∴ `dy/dx=square`
∴ equation (1) becomes
`x(dv)/dx = square`
∴ `square dv = dx/x`
On integrating, we get
`int(2v)/(1-v^2) dv =intdx/x`
∴ `-log|1-v^2|=log|x|+c_1`
∴ `log|x| + log|1-v^2|=logc ...["where" - c_1 = log c]`
∴ x(1 - v2) = c
By putting the value of v, the general solution of the D.E. is `square`= cx
Evaluate `int(1 + x + (x^2)/(2!))dx`
Evaluate the following.
`int (x^3)/(sqrt(1 + x^4))dx`
Evaluate:
`intcos^-1(sqrt(x))dx`
Evaluate:
`int e^(ax)*cos(bx + c)dx`
Evaluate the following.
`intx^3 e^(x^2) dx`
Evaluate the following.
`intx^3e^(x^2) dx`
If f′(x) = 4x3 − 3x2 + 2x + k, f(0) = 1 and f(1) = 4, find f(x).
The value of `inta^x.e^x dx` equals
Evaluate `int(1 + x + x^2/(2!))dx`.
Evaluate.
`int(5x^2 - 6x + 3)/(2x - 3) dx`
The value of `int (x sin^-1)/(sqrt(1 - x^2)) dx` is equal to:
`∫ sin^(−1)` xdx is equal to ______.
Repeated parts may be needed for which pair of integrals?
