Advertisements
Advertisements
प्रश्न
Write the primitive or anti-derivative of
\[f\left( x \right) = \sqrt{x} + \frac{1}{\sqrt{x}} .\]
बेरीज
उत्तर
` f (x) = \sqrtx + 1/ \sqrtx `.
integrating both sides
\[\int{f}\left( x \right)dx = \int\left( \sqrt{x} + \frac{1}{\sqrt{x}} \right)dx\]
\[ = \int\left( x^\frac{1}{2} + x^{- \frac{1}{2}} \right)dx\]
\[ = \left[ \frac{x^\frac{1}{2} + 1}{\frac{1}{2} + 1} \right] + \left[ \frac{x^{- \frac{1}{2} + 1}}{- \frac{1}{2} + 1} \right] + C\]
\[ = \frac{2}{3} x^\frac{3}{2} + 2 x^\frac{1}{2} + C\]
shaalaa.com
या प्रश्नात किंवा उत्तरात काही त्रुटी आहे का?
APPEARS IN
संबंधित प्रश्न
If f' (x) = x + b, f(1) = 5, f(2) = 13, find f(x)
\[\int\frac{1}{\sqrt{x + 1} + \sqrt{x}} dx\]
\[\int\text{sin mx }\text{cos nx dx m }\neq n\]
\[\int\frac{x + 1}{x \left( x + \log x \right)} dx\]
\[\int\left\{ 1 + \tan x \tan \left( x + \theta \right) \right\} dx\]
\[\int\frac{\sin 2x}{\left( a + b \cos 2x \right)^2} dx\]
\[\int2x \sec^3 \left( x^2 + 3 \right) \tan \left( x^2 + 3 \right) dx\]
\[\int 5^{x + \tan^{- 1} x} . \left( \frac{x^2 + 2}{x^2 + 1} \right) dx\]
\[\int\frac{\sec^2 \sqrt{x}}{\sqrt{x}} dx\]
\[\int\left( 2 x^2 + 3 \right) \sqrt{x + 2} \text{ dx }\]
` ∫ sec^6 x tan x dx `
\[\int {cosec}^4 \text{ 3x } \text{ dx } \]
\[\int \cot^6 x \text{ dx }\]
\[\int \cos^5 x \text{ dx }\]
\[\int\frac{1}{\sqrt{\left( 2 - x \right)^2 + 1}} dx\]
\[\int\frac{1}{\sqrt{\left( x - \alpha \right)\left( \beta - x \right)}} dx, \left( \beta > \alpha \right)\]
\[\int\frac{\cos 2x}{\sqrt{\sin^2 2x + 8}} dx\]
\[\int\frac{\cos x}{\sqrt{4 - \sin^2 x}} dx\]
` ∫ {x-3} /{ x^2 + 2x - 4 } dx `
\[\int\frac{x + 1}{\sqrt{x^2 + 1}} dx\]
\[\int\frac{\cos x}{\cos 3x} \text{ dx }\]
\[\int\frac{1}{5 + 4 \cos x} dx\]
\[\int x e^x \text{ dx }\]
\[\int\left( x + 1 \right) \text{ e}^x \text{ log } \left( x e^x \right) dx\]
\[\int \tan^{- 1} \left( \frac{2x}{1 - x^2} \right) \text{ dx }\]
\[\int \tan^{- 1} \left( \sqrt{x} \right) \text{dx }\]
\[\int\left( 2x - 5 \right) \sqrt{2 + 3x - x^2} \text{ dx }\]
\[\int\frac{5x}{\left( x + 1 \right) \left( x^2 - 4 \right)} dx\]
\[\int\frac{\sin^6 x}{\cos^8 x} dx =\]
\[\int \sin^3 x \cos^4 x\ \text{ dx }\]
\[\int \sin^5 x\ dx\]
\[\int\frac{1}{x^2 + 4x - 5} \text{ dx }\]
\[\int \left( x + 1 \right)^2 e^x \text{ dx }\]
\[\int\frac{x^2}{\left( x - 1 \right)^3 \left( x + 1 \right)} \text{ dx}\]
\[\int\frac{\cot x + \cot^3 x}{1 + \cot^3 x} \text{ dx}\]
Find : \[\int\frac{dx}{\sqrt{3 - 2x - x^2}}\] .
\[\int \left( e^x + 1 \right)^2 e^x dx\]
\[\int\frac{x^2}{x^2 + 7x + 10}\text{ dx }\]
\[\int\frac{3x + 1}{\sqrt{5 - 2x - x^2}} \text{ dx }\]
