Advertisements
Advertisements
Question
Integrate the function `(x + 2)/sqrt(x^2 -1)`
Advertisements
Solution
Let `I = (x + 2)/sqrt(x^2 - 1) dx`
`= int x/sqrt(x^2 - 1) dx + int 2/sqrt(x^2 - 1) dx`
`= I_1 + I_2 + C` (Say) ....(i)
Now `I_1 = int x/sqrt(x^2 - 1) dx`
Put x2 - 1 = t
2x dx = dt
`= 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` ....(ii)
and `I_2 = int 2/sqrt(x^2 - 1) dx`
`= 2 log abs (x + sqrt(x^2 - 1) + C_2` .....(iii)
From (i), (ii) and (iii), we get,
`therefore I = sqrt(x^2 - 1) + 2 log abs (x + sqrt(x^2 - 1)) + C`
APPEARS IN
RELATED QUESTIONS
Evaluate : ` int x^2/((x^2+4)(x^2+9))dx`
Evaluate:
`int((x+3)e^x)/((x+5)^3)dx`
Integrate the function `1/sqrt(1+4x^2)`
Integrate the function `1/sqrt((2-x)^2 + 1)`
Integrate the function `1/sqrt(9 - 25x^2)`
Integrate the function `x^2/(1 - x^6)`
Integrate the function `(x - 1)/sqrt(x^2 - 1)`
Integrate the function `x^2/sqrt(x^6 + a^6)`
Integrate the function `1/(9x^2 + 6x + 5)`
Integrate the function `1/sqrt((x - a)(x - b))`
Integrate the function `(6x + 7)/sqrt((x - 5)(x - 4))`
Integrate the function `(x+2)/sqrt(x^2 + 2x + 3)`
Integrate the function `(5x + 3)/sqrt(x^2 + 4x + 10)`
`int dx/(x^2 + 2x + 2)` equals:
`int dx/sqrt(9x - 4x^2)` equals:
Integrate the function:
`sqrt(4 - x^2)`
Integrate the function:
`sqrt(x^2 + 4x + 6)`
Integrate the function:
`sqrt(1+ 3x - x^2)`
Integrate the function:
`sqrt(x^2 + 3x)`
`int sqrt(1+ x^2) dx` is equal to ______.
Evaluate : `int_2^3 3^x dx`
Find `int (2x)/(x^2 + 1)(x^2 + 2)^2 dx`
\[\int\frac{8x + 13}{\sqrt{4x + 7}} \text{ dx }\]
\[\int\frac{1 + x + x^2}{x^2 \left( 1 + x \right)} \text{ dx}\]
Find:
`int_(-pi/4)^0 (1+tan"x")/(1-tan"x") "dx"`
Find `int (dx)/sqrt(4x - x^2)`
