Advertisements
Advertisements
प्रश्न
\[\int \left( \sqrt{x} - \frac{1}{\sqrt{x}} \right)^2 dx\]
बेरीज
उत्तर
\[\int \left( \sqrt{x} - \frac{1}{\sqrt{x}} \right)^2 dx\]
\[ = \int\left( x + \frac{1}{x} - 2 \right)dx\]
` = ∫ x dx + ∫ dx/x - 2\ ∫ x dx`
`= x^2 / 2 + In | x| -2 x +C`
shaalaa.com
या प्रश्नात किंवा उत्तरात काही त्रुटी आहे का?
APPEARS IN
संबंधित प्रश्न
\[\int\frac{\sin^2 x}{1 + \cos x} \text{dx} \]
` ∫ {cosec x} / {"cosec x "- cot x} ` dx
\[\int\frac{1}{1 - \sin\frac{x}{2}} dx\]
\[\int\frac{x^2 + 5x + 2}{x + 2} dx\]
\[\int\frac{2x - 1}{\left( x - 1 \right)^2} dx\]
\[\int\left( 5x + 3 \right) \sqrt{2x - 1} dx\]
\[\int\frac{e^{3x}}{e^{3x} + 1} dx\]
\[\int\frac{x}{\sqrt{x^2 + a^2} + \sqrt{x^2 - a^2}} dx\]
\[\int\sqrt {e^x- 1} \text{dx}\]
\[\int\frac{x^2}{\sqrt{x - 1}} dx\]
\[\int\left( 2 x^2 + 3 \right) \sqrt{x + 2} \text{ dx }\]
\[\int\frac{x^2 + 3x + 1}{\left( x + 1 \right)^2} dx\]
\[\int \sec^4 2x \text{ dx }\]
` ∫ {1}/{a^2 x^2- b^2}dx`
\[\int\frac{1}{x^{2/3} \sqrt{x^{2/3} - 4}} dx\]
\[\int\frac{x^2}{x^2 + 7x + 10} dx\]
\[\int\frac{6x - 5}{\sqrt{3 x^2 - 5x + 1}} \text{ dx }\]
\[\int\frac{1}{3 + 2 \cos^2 x} \text{ dx }\]
\[\int\frac{1}{\cos x \left( \sin x + 2 \cos x \right)} dx\]
\[\int x^2 e^{- x} \text{ dx }\]
` ∫ x tan ^2 x dx
\[\int \sin^3 \sqrt{x}\ dx\]
\[\int x \cos^3 x\ dx\]
\[\int\sqrt{2x - x^2} \text{ dx}\]
\[\int\left( x + 1 \right) \sqrt{x^2 - x + 1} \text{ dx}\]
Evaluate the following integral:
\[\int\frac{x^2}{\left( x^2 + a^2 \right)\left( x^2 + b^2 \right)}dx\]
\[\int\frac{1}{\left( 1 + x^2 \right) \sqrt{1 - x^2}} \text{ dx }\]
\[\int\frac{1}{\cos x + \sqrt{3} \sin x} \text{ dx } \] is equal to
\[\int\frac{e^x \left( 1 + x \right)}{\cos^2 \left( x e^x \right)} dx =\]
\[\int \sin^4 2x\ dx\]
\[\int \cot^5 x\ dx\]
\[\int\frac{x^3}{\left( 1 + x^2 \right)^2} \text{ dx }\]
\[\int\frac{1}{\left( \sin x - 2 \cos x \right) \left( 2 \sin x + \cos x \right)} \text{ dx }\]
\[\int\frac{1}{2 - 3 \cos 2x} \text{ dx }\]
\[\int\frac{\sin^5 x}{\cos^4 x} \text{ dx }\]
\[\int\frac{x^2}{\sqrt{1 - x}} \text{ dx }\]
Evaluate : \[\int\frac{\cos 2x + 2 \sin^2 x}{\cos^2 x}dx\] .
\[\int\frac{\cos^7 x}{\sin x} dx\]
Find: `int (sin2x)/sqrt(9 - cos^4x) dx`