Advertisements
Advertisements
प्रश्न
\[\int\frac{1}{a^2 - b^2 x^2} dx\]
बेरीज
उत्तर
\[\int\frac{dx}{a^2 - b^2 x^2}\]
\[ = \frac{1}{b^2}\int\frac{dx}{\left( \frac{a^2}{b^2} \right) - x^2} \]
\[ = \frac{1}{b^2} \times \frac{1}{2\frac{a}{b}} \log \left| \frac{\frac{a}{b} + x}{\frac{a}{b} - x} \right| + C \left[ \therefore \int\frac{dx}{a^2 - x^2} = \frac{1}{2a} \log \left| \frac{a + x}{a - x} \right| + C \right]\]
` = \text{1}/{2ab} \text{ log }|{a + bx}/{a - bx}| + c `
shaalaa.com
या प्रश्नात किंवा उत्तरात काही त्रुटी आहे का?
APPEARS IN
संबंधित प्रश्न
\[\int\left( \frac{m}{x} + \frac{x}{m} + m^x + x^m + mx \right) dx\]
\[\int\frac{\left( 1 + \sqrt{x} \right)^2}{\sqrt{x}} dx\]
\[\int\frac{\left( x + 1 \right)\left( x - 2 \right)}{\sqrt{x}} dx\]
\[\int\frac{\cos x}{1 - \cos x} \text{dx }or \int\frac{\cot x}{\text{cosec } {x }- \cot x} dx\]
\[\int \cos^{- 1} \left( \sin x \right) dx\]
\[\int\frac{\left( x^3 + 8 \right)\left( x - 1 \right)}{x^2 - 2x + 4} dx\]
\[\int\frac{1}{\sqrt{x + 1} + \sqrt{x}} dx\]
\[\int\frac{x + 1}{\sqrt{2x + 3}} dx\]
` ∫ sin x \sqrt (1-cos 2x) dx `
\[\int\frac{1 + \cot x}{x + \log \sin x} dx\]
\[\int\frac{\cos^3 x}{\sqrt{\sin x}} dx\]
\[\int\frac{x \sin^{- 1} x^2}{\sqrt{1 - x^4}} dx\]
\[\int\frac{\sin\sqrt{x}}{\sqrt{x}} dx\]
\[\int\frac{\sin \left( \text{log x} \right)}{x} dx\]
` ∫ x {tan^{- 1} x^2}/{1 + x^4} dx`
\[\int\left( 2 x^2 + 3 \right) \sqrt{x + 2} \text{ dx }\]
\[\int\frac{x^2 - 1}{x^2 + 4} dx\]
\[\int\frac{1}{\sqrt{\left( 1 - x^2 \right)\left\{ 9 + \left( \sin^{- 1} x \right)^2 \right\}}} dx\]
\[\int\frac{x^2 + x + 1}{x^2 - x} dx\]
\[\int\frac{x^2 + 1}{x^2 - 5x + 6} dx\]
\[\int\frac{2x + 5}{\sqrt{x^2 + 2x + 5}} dx\]
\[\int x\sqrt{x^2 + x} \text{ dx }\]
\[\int\frac{x^2 + x - 1}{x^2 + x - 6} dx\]
\[\int\frac{2x - 3}{\left( x^2 - 1 \right) \left( 2x + 3 \right)} dx\]
\[\int\frac{x^2 + x - 1}{\left( x + 1 \right)^2 \left( x + 2 \right)} dx\]
\[\int\frac{2x + 1}{\left( x - 2 \right) \left( x - 3 \right)} dx\]
\[\int\frac{1}{x^4 + x^2 + 1} \text{ dx }\]
\[\int\frac{\cos 2x - 1}{\cos 2x + 1} dx =\]
\[\int \cos^3 (3x)\ dx\]
\[\int\frac{1}{e^x + 1} \text{ dx }\]
\[\int\frac{\sin x}{\sqrt{1 + \sin x}} dx\]
\[\int\sqrt{\sin x} \cos^3 x\ \text{ dx }\]
\[\int\sqrt{\frac{1 - x}{x}} \text{ dx}\]
\[\int\frac{1}{\sin^2 x + \sin 2x} \text{ dx }\]
\[\int\frac{x^3}{\sqrt{x^8 + 4}} \text{ dx }\]
\[\int\frac{\cos x}{\frac{1}{4} - \cos^2 x} \text{ dx }\]
\[\int \sin^{- 1} \sqrt{x}\ dx\]
\[\int\frac{1}{\left( x^2 + 2 \right) \left( x^2 + 5 \right)} \text{ dx}\]
\[\int\sqrt{\frac{1 - \sqrt{x}}{1 + \sqrt{x}}} \text{ dx}\]
Find: `int (sin2x)/sqrt(9 - cos^4x) dx`
