Advertisements
Advertisements
Question
Integrate the function `(x - 1)/sqrt(x^2 - 1)`
Solution
Let `I = int (x - 1)/sqrt(x^2 - 1) dx`
`= int x/(sqrt(x^2 - 1)) dx - int 1/sqrt(x^2 - 1) dx`
`= I_1 - I_2` (say)
Now , `I_1 = x/sqrt(x^2 - 1) dx`
Put x2 - 1 = t
2x dx = dt ⇒ x dx = `1/2` dt
`therefore I = 1/2 int dt/sqrtt = 1/2 int t^((-1)/2) dt`
`= 1/2 xx t^(1/2)/(1/2) + C_1 = sqrtt = C_1`
`= sqrt(x^2 - 1) + C_1`
and `I_2 int 1/sqrt(x^2 - 1) dx`
`= log [x + sqrt(x^2) - 1] + C_2` `....[∵ int dx/sqrt(x^2 - a^2) = log |x + sqrt(x^2 - a^2)| + C]`
`therefore I = sqrt(x^2 - 1) - log |x + sqrt(x^2 - 1)| +C`
APPEARS IN
RELATED QUESTIONS
Evaluate : ` int x^2/((x^2+4)(x^2+9))dx`
Find:
`int(x^3-1)/(x^3+x)dx`
Evaluate:
`int((x+3)e^x)/((x+5)^3)dx`
Integrate the function `1/sqrt(1+4x^2)`
Integrate the function `1/sqrt(9 - 25x^2)`
Integrate the function `x^2/sqrt(x^6 + a^6)`
Integrate the function `(sec^2 x)/sqrt(tan^2 x + 4)`
Integrate the function `1/sqrt(x^2 +2x + 2)`
Integrate the function `1/(9x^2 + 6x + 5)`
Integrate the function `1/sqrt((x -1)(x - 2))`
Integrate the function `1/sqrt(8+3x - x^2)`
Integrate the function `(5x - 2)/(1 + 2x + 3x^2)`
Integrate the function `(6x + 7)/sqrt((x - 5)(x - 4))`
Integrate the function `(x+2)/sqrt(x^2 + 2x + 3)`
Integrate the function `(x + 3)/(x^2 - 2x - 5)`
Integrate the function `(5x + 3)/sqrt(x^2 + 4x + 10)`
Integrate the function:
`sqrt(1-4x - x^2)`
Integrate the function:
`sqrt(1+ 3x - x^2)`
`int sqrt(1+ x^2) dx` is equal to ______.
Find `int (2x)/(x^2 + 1)(x^2 + 2)^2 dx`
Integration of \[\frac{1}{1 + \left( \log_e x \right)^2}\] with respect to loge x is
\[\int\frac{1 + x + x^2}{x^2 \left( 1 + x \right)} \text{ dx}\]
Find:
`int_(-pi/4)^0 (1+tan"x")/(1-tan"x") "dx"`
If θ f(x) = `int_0^x t sin t dt` then `f^1(x)` is
Find `int (dx)/sqrt(4x - x^2)`
Find: `int (dx)/(x^2 - 6x + 13)`
`int (a^x - b^x)^2/(a^xb^x)dx` equals ______.
