Advertisements
Advertisements
Question
Evaluate the following.
`int 1/(sqrt"x" + "x")` dx
Advertisements
Solution
Let I = `int 1/(sqrt"x" + "x")` dx
= `int 1/(sqrt"x" (1 + sqrt"x"))`dx
Put `1 + sqrt"x" = "t"`
∴ `1/(2sqrt"x") "dx" = "dt"`
∴ `1/sqrt"x"`dx = 2 dt
∴ I = `int (2 * "dt")/"t"`
`= 2 int 1/"t" * "dt"`
= 2 log | t | + c
∴ I = 2 log `|1 + sqrt"x"|` + c
APPEARS IN
RELATED QUESTIONS
Evaluate :
`∫(x+2)/sqrt(x^2+5x+6)dx`
Evaluate: `int sqrt(tanx)/(sinxcosx) dx`
Integrate the functions:
`(sin x)/(1+ cos x)^2`
Evaluate: `int_0^3 f(x)dx` where f(x) = `{(cos 2x, 0<= x <= pi/2),(3, pi/2 <= x <= 3) :}`
Write a value of
Evaluate: \[\int\frac{x^3 - 1}{x^2} \text{ dx}\]
Evaluate : `int ("e"^"x" (1 + "x"))/("cos"^2("x""e"^"x"))"dx"`
Evaluate the following integrals: `int(x - 2)/sqrt(x + 5).dx`
Integrate the following functions w.r.t. x : `(sinx + 2cosx)/(3sinx + 4cosx)`
Evaluate the following : `int (1)/(4x^2 - 3).dx`
Evaluate the following : `int (1)/(cos2x + 3sin^2x).dx`
Integrate the following functions w.r.t. x : `int (1)/(2sin 2x - 3)dx`
Evaluate the following.
`int "x" sqrt(1 + "x"^2)` dx
Choose the correct alternative from the following.
The value of `int "dx"/sqrt"1 - x"` is
`int ("e"^(3x))/("e"^(3x) + 1) "d"x`
To find the value of `int ((1 + logx))/x` dx the proper substitution is ______
`int1/(4 + 3cos^2x)dx` = ______
`int(3x + 1)/(2x^2 - 2x + 3)dx` equals ______.
`int (x + sinx)/(1 + cosx)dx` is equal to ______.
The value of `sqrt(2) int (sinx dx)/(sin(x - π/4))` is ______.
`int (cos4x)/(sin2x + cos2x)dx` = ______.
Evaluate the following.
`intx^3/sqrt(1+x^4)dx`
Evaluate `int (5x^2 - 6x + 3)/(2x - 3) dx`
If f'(x) = 4x3 – 3x2 + 2x + k, f(0) = 1 and f(1) = 4, find f(x).
Evaluate:
`intsqrt(sec x/2 - 1)dx`
