Advertisements
Advertisements
प्रश्न
\[\int\frac{\left( 1 + \sqrt{x} \right)^2}{\sqrt{x}} dx\]
योग
Advertisements
उत्तर
\[\int \frac{\left( 1 + \sqrt{x} \right)^2}{\sqrt{x}}dx\]
\[ = \int \left( \frac{1}{\sqrt{x}} + \frac{x}{\sqrt{x}} + 2\frac{\sqrt{x}}{\sqrt{x}} \right)dx\]
\[ = \int\left( x^{- \frac{1}{2}} + x^\frac{1}{2} + 2 \right)dx\]
\[ = \left[ \frac{x^{- \frac{1}{2} + 1}}{- \frac{1}{2} + 1} \right] + \left[ \frac{x^\frac{1}{2} + 1}{\frac{1}{2} + 1} \right] + 2x + C\]
\[ = 2\sqrt{x} + \frac{2}{3} x^\frac{3}{2} + 2x + C\]
shaalaa.com
क्या इस प्रश्न या उत्तर में कोई त्रुटि है?
APPEARS IN
संबंधित प्रश्न
`int{sqrtx(ax^2+bx+c)}dx`
\[\int\left( 2 - 3x \right) \left( 3 + 2x \right) \left( 1 - 2x \right) dx\]
\[\int \left( \sqrt{x} - \frac{1}{\sqrt{x}} \right)^2 dx\]
\[\int\frac{\cos x}{1 - \cos x} \text{dx }or \int\frac{\cot x}{\text{cosec } {x }- \cot x} dx\]
\[\int\frac{1}{\sqrt{2x + 3} + \sqrt{2x - 3}} dx\]
` ∫ cos mx cos nx dx `
\[\int\frac{\text{sin} \left( x - \alpha \right)}{\text{sin }\left( x + \alpha \right)} dx\]
\[\int\frac{1}{x (3 + \log x)} dx\]
\[\int\frac{a}{b + c e^x} dx\]
\[\int\frac{\cos\sqrt{x}}{\sqrt{x}} dx\]
` ∫ tan^5 x sec ^4 x dx `
\[\int \cot^n {cosec}^2 \text{ x dx } , n \neq - 1\]
\[\int\frac{1}{a^2 x^2 + b^2} dx\]
\[\int\frac{1}{\sqrt{a^2 - b^2 x^2}} dx\]
\[\int\frac{1}{x^2 + 6x + 13} dx\]
\[\int\frac{e^x}{e^{2x} + 5 e^x + 6} dx\]
\[\int x \cos^2 x\ dx\]
\[\int x^2 \sin^{- 1} x\ dx\]
\[\int\left( \tan^{- 1} x^2 \right) x\ dx\]
\[\int e^x \left( \frac{1}{x^2} - \frac{2}{x^3} \right) dx\]
\[\int e^x \left[ \sec x + \log \left( \sec x + \tan x \right) \right] dx\]
\[\int e^x \left( \log x + \frac{1}{x} \right) dx\]
∴\[\int e^{2x} \left( - \sin x + 2 \cos x \right) dx\]
\[\int\sqrt{3 - 2x - 2 x^2} \text{ dx}\]
\[\int\frac{x^2 + x - 1}{x^2 + x - 6} dx\]
\[\int\frac{x^2}{\left( x - 1 \right) \left( x - 2 \right) \left( x - 3 \right)} dx\]
\[\int\frac{2x + 1}{\left( x - 2 \right) \left( x - 3 \right)} dx\]
\[\int\frac{x^2 + 9}{x^4 + 81} \text{ dx }\]
\[\int \sec^2 x \cos^2 2x \text{ dx }\]
\[\int\frac{1}{1 + 2 \cos x} \text{ dx }\]
\[\int\frac{1 + \sin x}{\sin x \left( 1 + \cos x \right)} \text{ dx }\]
\[\int\frac{\sin^6 x}{\cos x} \text{ dx }\]
\[\int\sqrt{3 x^2 + 4x + 1}\text{ dx }\]
\[\int \left( x + 1 \right)^2 e^x \text{ dx }\]
\[\int \sin^{- 1} \sqrt{\frac{x}{a + x}} \text{ dx}\]
\[\int\frac{x^2}{\left( x - 1 \right)^3 \left( x + 1 \right)} \text{ dx}\]
\[\int\frac{x^2 - 2}{x^5 - x} \text{ dx}\]
\[\int\frac{x^2 + x + 1}{\left( x + 1 \right)^2 \left( x + 2 \right)} \text{ dx}\]
\[\int \sin^3 \left( 2x + 1 \right) \text{dx}\]
Find: `int (3x +5)/(x^2+3x-18)dx.`
