Advertisements
Advertisements
Question
`int sqrt(1 + "x"^2) "dx"` =
Options
`"x"/2 sqrt(1 + "x"^2) + 1/2 log ("x" + sqrt(1 + "x"^2))`+ c
`2/3 (1 + "x"^2)^(3/2) + "c"`
`1/3 (1 + "x"^2)` + c
`("(x)")/sqrt(1 + "x"^2)` + c
Advertisements
Solution
`"x"/2 sqrt(1 + "x"^2) + 1/2 log ("x" + sqrt(1 + "x"^2))`+ c
Explanation:
∵ `int sqrt(a^2 + "x"^2) "dx"` = `x/2 sqrt(a^2 + x^2) + a^2/2 log | x + sqrt(a^2 + x^2)| + c`
∴ I = `x/2 sqrt(1 + x^2) + 1/2 log |x + sqrt(1 + x^2) + c`
APPEARS IN
RELATED QUESTIONS
Integrate the functions:
`x/(sqrt(x+ 4))`, x > 0
Integrate the functions:
`1/(x(log x)^m), x > 0, m ne 1`
Integrate the functions:
`1/(cos^2 x(1-tan x)^2`
`(10x^9 + 10^x log_e 10)/(x^10 + 10^x) dx` equals:
Write a value of
Write a value of\[\int\text{ tan x }\sec^3 x\ dx\]
Write a value of
Evaluate: \[\int\frac{x^3 - 1}{x^2} \text{ dx}\]
Integrate the following function w.r.t. x:
x9.sec2(x10)
Integrate the following functions w.r.t. x : cos7x
Evaluate the following : `int (1)/sqrt(11 - 4x^2).dx`
Evaluate the following : `int sqrt((9 + x)/(9 - x)).dx`
Integrate the following functions w.r.t. x : `int (1)/(3 - 2cos 2x).dx`
Choose the correct options from the given alternatives :
`int f x^x (1 + log x)*dx`
Evaluate `int 1/(x (x - 1))` dx
Evaluate the following.
`int (3"e"^"x" + 4)/(2"e"^"x" - 8)`dx
Evaluate the following.
`int "x"^3/(16"x"^8 - 25)` dx
Fill in the Blank.
`int 1/"x"^3 [log "x"^"x"]^2 "dx" = "P" (log "x")^3` + c, then P = _______
Evaluate: `int log ("x"^2 + "x")` dx
Evaluate: `int sqrt("x"^2 + 2"x" + 5)` dx
`int "dx"/((sin x + cos x)(2 cos x + sin x))` = ?
`int(sin2x)/(5sin^2x+3cos^2x) dx=` ______.
`int cos^3x dx` = ______.
Evaluate `int(1 + x + x^2/(2!))dx`
Evaluate.
`int (5x^2 - 6x + 3)/(2x - 3) 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).
The value of `int ("d"x)/(sqrt(1 - x))` is ______.
Evaluate `int(1+x+x^2/(2!))dx`
