Advertisements
Advertisements
Question
Integrate the functions:
`e^(tan^(-1)x)/(1+x^2)`
Advertisements
Solution
Let `I = int (e^(tan -1 x)/(1 + x^2))` dx
Put tan-1 x = t
`1/(1 + x^2)` dx = dt
Hence, `I = int e^t` dt
= et + C
= `e^(tan^-1) + C`
APPEARS IN
RELATED QUESTIONS
Evaluate :
`int(sqrt(cotx)+sqrt(tanx))dx`
Find : `int((2x-5)e^(2x))/(2x-3)^3dx`
Integrate the functions:
`(log x)^2/x`
Integrate the functions:
sin (ax + b) cos (ax + b)
Integrate the functions:
`sqrt(ax + b)`
Integrate the functions:
`x/(sqrt(x+ 4))`, x > 0
Integrate the functions:
`(2cosx - 3sinx)/(6cos x + 4 sin x)`
Integrate the functions:
`((x+1)(x + logx)^2)/x`
Write a value of
Write a value of\[\int \log_e x\ dx\].
Prove that: `int "dx"/(sqrt("x"^2 +"a"^2)) = log |"x" +sqrt("x"^2 +"a"^2) | + "c"`
Integrate the following w.r.t. x : `int x^2(1 - 2/x)^2 dx`
If `f'(x) = x - (3)/x^3, f(1) = (11)/(2)`, find f(x)
Integrate the following functions w.r.t. x : `(2x + 1)sqrt(x + 2)`
Integrate the following functions w.r.t. x : `(4e^x - 25)/(2e^x - 5)`
Integrate the following functions w.r.t. x : tan 3x tan 2x tan x
Evaluate the following : `int sqrt((9 + x)/(9 - x)).dx`
Choose the correct option from the given alternatives :
`int (1 + x + sqrt(x + x^2))/(sqrt(x) + sqrt(1 + x))*dx` =
Choose the correct options from the given alternatives :
`int dx/(cosxsqrt(sin^2x - cos^2x))*dx` =
Evaluate the following.
`int 1/(x(x^6 + 1))` dx
Evaluate the following.
`int ((3"e")^"2t" + 5)/(4"e"^"2t" - 5)`dt
Evaluate: `int sqrt(x^2 - 8x + 7)` dx
`int ((x + 1)(x + log x))^4/(3x) "dx" =`______.
`int(sin2x)/(5sin^2x+3cos^2x) dx=` ______.
`int_1^3 ("d"x)/(x(1 + logx)^2)` = ______.
The value of `int (sinx + cosx)/sqrt(1 - sin2x) dx` is equal to ______.
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)
Evaluate `int (1+x+x^2/(2!)) dx`
Evaluate:
`int sqrt((a - x)/x) dx`
Evaluate:
`int(sqrt(tanx) + sqrt(cotx))dx`
Evaluate `int 1/(x(x-1))dx`
Evaluate the following:
`int x^3/(sqrt(1+x^4))dx`
Evaluate `int1/(x(x-1))dx`
If f '(x) = 4x3 - 3x2 + 2x + k, f(0) = 1 and f(1) = 4, find f(x).
What must be done before integrating after obtaining \[du\]?
In \[\int\sin^3x\cos^2x\,dx\], which rewriting prepares the integrand for the substitution \[t=\cos x\]?
Which expression results after simplifying \[\int\frac{\sin(t-a)}{\sin t}\,dt\]?
