Advertisements
Advertisements
Question
\[\int\frac{2x - 1}{\left( x - 1 \right)^2} dx\]
Sum
Advertisements
Solution
\[\int\left[ \frac{2x - 1}{\left( x - 1 \right)^2} \right]dx\]
\[\text{Let x - 1 }= t\]
\[ \Rightarrow x = t + 1\]
\[ \Rightarrow 1 = \frac{dt}{dx}\]
\[Now, \int\left[ \frac{2x - 1}{\left( x - 1 \right)^2} \right]dx\]
\[ = \int\left[ \frac{2\left( t + 1 \right) - t}{t^2} \right]\text{ dt }\]
\[ = \int\left( \frac{2t + 1}{t^2} \right)\text{ dt }\]
\[ = 2\int\frac{dt}{t} + \int t^{- 2} \text{ dt }\]
\[ = \text{ 2 log }\left| t \right| + \frac{t^{- 2 + 1}}{- 2 + 1} + C\]
\[ =\text{ 2 log }\left( x - 1 \right) - \frac{1}{x - 1} + C\]
\[\text{Let x - 1 }= t\]
\[ \Rightarrow x = t + 1\]
\[ \Rightarrow 1 = \frac{dt}{dx}\]
\[Now, \int\left[ \frac{2x - 1}{\left( x - 1 \right)^2} \right]dx\]
\[ = \int\left[ \frac{2\left( t + 1 \right) - t}{t^2} \right]\text{ dt }\]
\[ = \int\left( \frac{2t + 1}{t^2} \right)\text{ dt }\]
\[ = 2\int\frac{dt}{t} + \int t^{- 2} \text{ dt }\]
\[ = \text{ 2 log }\left| t \right| + \frac{t^{- 2 + 1}}{- 2 + 1} + C\]
\[ =\text{ 2 log }\left( x - 1 \right) - \frac{1}{x - 1} + C\]
shaalaa.com
Is there an error in this question or solution?
APPEARS IN
RELATED QUESTIONS
\[\int\frac{\left( 1 + \sqrt{x} \right)^2}{\sqrt{x}} dx\]
\[\int \left( 3x + 4 \right)^2 dx\]
\[\int\frac{\cos x}{1 - \cos x} \text{dx }or \int\frac{\cot x}{\text{cosec } {x }- \cot x} dx\]
\[\int\frac{1 + \cos x}{1 - \cos x} dx\]
\[\int\frac{1 + \cos 4x}{\cot x - \tan x} dx\]
\[\int\frac{x^2 + x + 5}{3x + 2} dx\]
\[\int \sin^2 \frac{x}{2} dx\]
\[\int\frac{1 - \cot x}{1 + \cot x} dx\]
\[\int\frac{1}{\sqrt{1 - x^2}\left( 2 + 3 \sin^{- 1} x \right)} dx\]
\[\int\frac{1 + \cot x}{x + \log \sin x} dx\]
\[\int\frac{x \sin^{- 1} x^2}{\sqrt{1 - x^4}} dx\]
` ∫ x {tan^{- 1} x^2}/{1 + x^4} dx`
` ∫ tan^5 x sec ^4 x dx `
\[\int \cot^6 x \text{ dx }\]
\[\int\frac{e^x}{\left( 1 + e^x \right)\left( 2 + e^x \right)} dx\]
\[\int\frac{1}{\sqrt{16 - 6x - x^2}} dx\]
\[\int\frac{x^2 \left( x^4 + 4 \right)}{x^2 + 4} \text{ dx }\]
\[\int\frac{x}{\sqrt{8 + x - x^2}} dx\]
\[\int\frac{1}{3 + 2 \cos^2 x} \text{ dx }\]
\[\int\frac{1}{5 + 4 \cos x} dx\]
\[\int\frac{x + \sin x}{1 + \cos x} \text{ dx }\]
\[\int x \sin x \cos 2x\ dx\]
\[\int\frac{x \sin^{- 1} x}{\sqrt{1 - x^2}} dx\]
\[\int e^x \left( \cot x - {cosec}^2 x \right) dx\]
\[\int\left( x + 2 \right) \sqrt{x^2 + x + 1} \text{ dx }\]
\[\int\frac{x^2 + x - 1}{x^2 + x - 6} dx\]
\[\int\frac{1}{x \log x \left( 2 + \log x \right)} dx\]
\[\int\frac{5 x^2 - 1}{x \left( x - 1 \right) \left( x + 1 \right)} dx\]
\[\int\frac{x^2 + x + 1}{\left( x + 1 \right)^2 \left( x + 2 \right)} dx\]
\[\int\frac{x}{\left( x^2 + 2x + 2 \right) \sqrt{x + 1}} \text{ dx}\]
If \[\int\frac{1}{\left( x + 2 \right)\left( x^2 + 1 \right)}dx = a\log\left| 1 + x^2 \right| + b \tan^{- 1} x + \frac{1}{5}\log\left| x + 2 \right| + C,\] then
\[\int \cot^5 x\ dx\]
\[\int\frac{1}{a + b \tan x} \text{ dx }\]
\[\int\frac{\sin^5 x}{\cos^4 x} \text{ dx }\]
\[\int\sqrt{1 + 2x - 3 x^2}\text{ dx } \]
\[ \int\left( 1 + x^2 \right) \ \cos 2x \ dx\]
\[\int \sin^{- 1} \sqrt{x}\ dx\]
\[\int\frac{x^2 + x + 1}{\left( x + 1 \right)^2 \left( x + 2 \right)} \text{ dx}\]
\[\int\frac{3x + 1}{\sqrt{5 - 2x - x^2}} \text{ dx }\]
