Advertisements
Advertisements
Question
Advertisements
Solution
\[\int\frac{dx}{\sqrt{\left( 2 - x \right)^2 - 1}}\]
\[\text{ let 2 }- x = t\]
\[ \Rightarrow - dx = dt\]
\[ \Rightarrow dx = - dt\]
\[Now, \int\frac{dx}{\sqrt{\left( 2 - x \right)^2 - 1}}\]
\[ = \int\frac{- dt}{\sqrt{t^2 - 1}}\]
\[ = - \text{ log }\left| t + \sqrt{t^2 - 1} \right| + C\]
\[ = - \text{ log }\left| \left( 2 - x \right) + \sqrt{\left( 2 - x \right)^2 - 1} \right| + C\]
APPEARS IN
RELATED QUESTIONS
If f' (x) = x + b, f(1) = 5, f(2) = 13, find f(x)
\[\int \tan^2 \left( 2x - 3 \right) dx\]
\[\int\frac{x^2 + 5x + 2}{x + 2} dx\]
` ∫ \sqrt{tan x} sec^4 x dx `
If \[\int\frac{\sin^8 x - \cos^8 x}{1 - 2 \sin^2 x \cos^2 x} dx\]
The primitive of the function \[f\left( x \right) = \left( 1 - \frac{1}{x^2} \right) a^{x + \frac{1}{x}} , a > 0\text{ is}\]
\[\int\frac{1}{4 \sin^2 x + 4 \sin x \cos x + 5 \cos^2 x} \text{ dx }\]
