Advertisements
Advertisements
प्रश्न
\[\int\sqrt{x^2 - 2x} \text{ dx}\]
योग
Advertisements
उत्तर
\[I = \int\sqrt{x^2 - 2x}dx\]
\[\Rightarrow I = \int\sqrt{x^2 - 2x + 1 - 1}\text{ dx}\]
\[ \Rightarrow I = \int\sqrt{(x - 1 )^2 - 1^2}dx\]
\[ \because \int\sqrt{x^2 - a^2}dx = \frac{x}{2}\sqrt{x^2 - a^2} - \frac{a^2}{2}\text{ ln}\left( \left| x + \sqrt{x^2 - a^2} \right| \right) + c\]
\[ \therefore I = \frac{(x - 1)}{2}\sqrt{(x - 1 )^2 - 1} - \frac{1}{2}\text{ ln}\left| \left( x - 1 \right) + \sqrt{x^2 - 2x} \right| + c\]
\[ \Rightarrow I = \int\sqrt{(x - 1 )^2 - 1^2}dx\]
\[ \because \int\sqrt{x^2 - a^2}dx = \frac{x}{2}\sqrt{x^2 - a^2} - \frac{a^2}{2}\text{ ln}\left( \left| x + \sqrt{x^2 - a^2} \right| \right) + c\]
\[ \therefore I = \frac{(x - 1)}{2}\sqrt{(x - 1 )^2 - 1} - \frac{1}{2}\text{ ln}\left| \left( x - 1 \right) + \sqrt{x^2 - 2x} \right| + c\]
shaalaa.com
क्या इस प्रश्न या उत्तर में कोई त्रुटि है?
APPEARS IN
संबंधित प्रश्न
\[\int \left( a \tan x + b \cot x \right)^2 dx\]
\[\int\frac{1}{2 - 3x} + \frac{1}{\sqrt{3x - 2}} dx\]
\[\int\frac{1}{\text{cos}^2\text{ x }\left( 1 - \text{tan x} \right)^2} dx\]
`∫ cos ^4 2x dx `
\[\int\frac{1}{\sqrt{1 - x^2}\left( 2 + 3 \sin^{- 1} x \right)} dx\]
` ∫ e^{m sin ^-1 x}/ \sqrt{1-x^2} ` dx
\[\int\frac{\cos\sqrt{x}}{\sqrt{x}} dx\]
\[\int\frac{1}{x^2 \left( x^4 + 1 \right)^{3/4}} dx\]
\[\int\frac{x^2}{\sqrt{3x + 4}} dx\]
\[\int\frac{1}{\sin^4 x \cos^2 x} dx\]
\[\int\frac{1}{\sqrt{\left( 2 - x \right)^2 - 1}} dx\]
\[\int\frac{x}{x^4 - x^2 + 1} dx\]
\[\int\frac{x}{\sqrt{x^4 + a^4}} dx\]
\[\int\frac{x}{\sqrt{4 - x^4}} dx\]
\[\int\frac{1}{4 \cos x - 1} \text{ dx }\]
\[\int\frac{1}{2 + \sin x + \cos x} \text{ dx }\]
\[\int\frac{1}{p + q \tan x} \text{ dx }\]
`int"x"^"n"."log" "x" "dx"`
\[\int {cosec}^3 x\ dx\]
\[\int x\left( \frac{\sec 2x - 1}{\sec 2x + 1} \right) dx\]
\[\int \sin^{- 1} \left( 3x - 4 x^3 \right) \text{ dx }\]
\[\int \cos^{- 1} \left( \frac{1 - x^2}{1 + x^2} \right) \text{ dx }\]
\[\int\left( \frac{1}{\log x} - \frac{1}{\left( \log x \right)^2} \right) dx\]
\[\int x\sqrt{x^4 + 1} \text{ dx}\]
\[\int\frac{x}{\left( x^2 - a^2 \right) \left( x^2 - b^2 \right)} dx\]
\[\int\frac{2 x^2 + 7x - 3}{x^2 \left( 2x + 1 \right)} dx\]
\[\int\frac{5 x^2 + 20x + 6}{x^3 + 2 x^2 + x} dx\]
\[\int\frac{18}{\left( x + 2 \right) \left( x^2 + 4 \right)} dx\]
Evaluate the following integral:
\[\int\frac{x^2}{1 - x^4}dx\]
\[\int\frac{1}{\left( x - 1 \right) \sqrt{x + 2}} \text{ dx }\]
\[\int e^x \left( \frac{1 - \sin x}{1 - \cos x} \right) dx\]
\[\int\frac{\sin x + \cos x}{\sqrt{\sin 2x}} \text{ dx}\]
\[\int \cot^4 x\ dx\]
\[\int \sin^5 x\ dx\]
\[\int\sqrt{3 x^2 + 4x + 1}\text{ dx }\]
\[\int x\sqrt{1 + x - x^2}\text{ dx }\]
\[\int \sin^{- 1} \sqrt{x}\ dx\]
\[\int e^x \frac{\left( 1 - x \right)^2}{\left( 1 + x^2 \right)^2} \text{ dx }\]
\[\int \left( e^x + 1 \right)^2 e^x dx\]
