Advertisements
Advertisements
प्रश्न
\[\int\frac{2x - 1}{\left( x - 1 \right)^2} dx\]
योग
Advertisements
उत्तर
\[\int\left[ \frac{2x - 1}{\left( x - 1 \right)^2} \right]dx\]
\[ = \int\left[ \frac{2x - 2 + 2 - 1}{\left( x - 1 \right)^2} \right]dx\]
\[ = \int\left( \frac{2 \left( x - 1 \right)}{\left( x - 1 \right)^2} + \frac{1}{\left( x - 1 \right)^2} \right)dx\]
\[ = 2\int\frac{dx}{x - 1} + \int \left( x - 1 \right)^{- 2} dx\]
\[ = \text{2 ln }\left| x - 1 \right| + \frac{\left( x - 1 \right)^{- 2 + 1}}{- 2 + 1} + C\]
\[ = \text{2 ln }\left| x - 1 \right| - \frac{1}{x - 1} + C\]
shaalaa.com
क्या इस प्रश्न या उत्तर में कोई त्रुटि है?
APPEARS IN
संबंधित प्रश्न
\[\int\frac{\left( 1 + x \right)^3}{\sqrt{x}} dx\]
\[\int\left( \sec^2 x + {cosec}^2 x \right) dx\]
\[\int\frac{1}{1 - \cos x} dx\]
\[\int\frac{1}{1 - \cos 2x} dx\]
\[\int\frac{\cos x}{1 + \cos x} dx\]
\[\int\frac{1 - \cos x}{1 + \cos x} dx\]
If f' (x) = 8x3 − 2x, f(2) = 8, find f(x)
\[\int\frac{x + 3}{\left( x + 1 \right)^4} dx\]
\[\int\frac{1}{1 - \sin\frac{x}{2}} dx\]
\[\int \left( e^x + 1 \right)^2 e^x dx\]
\[\int\frac{x^3}{x - 2} dx\]
\[\int\frac{3x + 5}{\sqrt{7x + 9}} dx\]
\[\int \sin^2 \frac{x}{2} dx\]
` ∫ sin x \sqrt (1-cos 2x) dx `
\[\int\frac{e^{3x}}{e^{3x} + 1} dx\]
\[\int\frac{\cos^3 x}{\sqrt{\sin x}} dx\]
\[\int \tan^{3/2} x \sec^2 \text{x dx}\]
\[\int\frac{e^{2x}}{1 + e^x} dx\]
\[\int\frac{x^2 + 3x + 1}{\left( x + 1 \right)^2} dx\]
\[\int \cot^6 x \text{ dx }\]
\[\int \sin^3 x \cos^6 x \text{ dx }\]
Evaluate the following integrals:
\[\int\frac{x^2}{\left( a^2 - x^2 \right)^{3/2}}dx\]
\[\int\frac{1}{x^2 - 10x + 34} dx\]
\[\int\frac{\cos x}{\sqrt{\sin^2 x - 2 \sin x - 3}} dx\]
\[\int\frac{x^2 + x - 1}{x^2 + x - 6}\text{ dx }\]
\[\int\frac{2x + 1}{\sqrt{x^2 + 2x - 1}}\text{ dx }\]
\[\int\frac{x}{\sqrt{x^2 + x + 1}} \text{ dx }\]
\[\int\cos\sqrt{x}\ dx\]
\[\int x \cos^3 x\ dx\]
\[\int\frac{1}{1 + x + x^2 + x^3} dx\]
\[\int\frac{dx}{\left( x^2 + 1 \right) \left( x^2 + 4 \right)}\]
\[\int\frac{3}{\left( 1 - x \right) \left( 1 + x^2 \right)} dx\]
\[\int\frac{x^2 + 1}{x^4 + 7 x^2 + 1} 2 \text{ dx }\]
\[\int\left( x - 1 \right) e^{- x} dx\] is equal to
\[\int\frac{x^9}{\left( 4 x^2 + 1 \right)^6}dx\] is equal to
\[\int\frac{x^3}{x + 1}dx\] is equal to
\[\int x\sqrt{\frac{1 - x}{1 + x}} \text{ dx }\]
\[\int \cos^{- 1} \left( 1 - 2 x^2 \right) \text{ dx }\]
