Advertisements
Advertisements
प्रश्न
Write the primitive or anti-derivative of
\[f\left( x \right) = \sqrt{x} + \frac{1}{\sqrt{x}} .\]
योग
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
क्या इस प्रश्न या उत्तर में कोई त्रुटि है?
APPEARS IN
संबंधित प्रश्न
\[\int\left( \frac{m}{x} + \frac{x}{m} + m^x + x^m + mx \right) dx\]
\[\int\left( x^e + e^x + e^e \right) dx\]
\[\int\frac{\cos x}{1 - \cos x} \text{dx }or \int\frac{\cot x}{\text{cosec } {x }- \cot x} dx\]
\[\int\frac{1}{\sqrt{x + 1} + \sqrt{x}} dx\]
\[\int\frac{x^2 + x + 5}{3x + 2} dx\]
\[\int\frac{2x - 1}{\left( x - 1 \right)^2} dx\]
\[\int\frac{2x + 1}{\sqrt{3x + 2}} dx\]
\[\int\frac{\sec^2 x}{\tan x + 2} dx\]
\[\int2x \sec^3 \left( x^2 + 3 \right) \tan \left( x^2 + 3 \right) dx\]
\[\int \tan^3 \text{2x sec 2x dx}\]
\[\int\sqrt {e^x- 1} \text{dx}\]
` ∫ tan^5 x sec ^4 x dx `
\[\int \cos^5 x \text{ dx }\]
\[\int\frac{1}{2 x^2 - x - 1} dx\]
\[\int\frac{1}{\sqrt{7 - 6x - x^2}} dx\]
\[\int\frac{\sin x}{\sqrt{4 \cos^2 x - 1}} dx\]
` ∫ {x-3} /{ x^2 + 2x - 4 } dx `
\[\int\frac{x + 7}{3 x^2 + 25x + 28}\text{ dx}\]
\[\int\frac{\left( x - 1 \right)^2}{x^2 + 2x + 2} dx\]
\[\int\frac{1}{\sin^2 x + \sin 2x} \text{ dx }\]
\[\int x^2 e^{- x} \text{ dx }\]
\[\int x^2 \sin^2 x\ dx\]
` ∫ sin x log (\text{ cos x ) } dx `
\[\int e^x \left( \frac{\sin 4x - 4}{1 - \cos 4x} \right) dx\]
\[\int\frac{x^2 + 6x - 8}{x^3 - 4x} dx\]
\[\int\frac{1}{x\left[ 6 \left( \log x \right)^2 + 7 \log x + 2 \right]} dx\]
\[\int\frac{x^2 + 1}{\left( x - 2 \right)^2 \left( x + 3 \right)} dx\]
\[\int\frac{x^2 + x - 1}{\left( x + 1 \right)^2 \left( x + 2 \right)} dx\]
\[\int\frac{18}{\left( x + 2 \right) \left( x^2 + 4 \right)} dx\]
Evaluate the following integral:
\[\int\frac{x^2}{1 - x^4}dx\]
If \[\int\frac{1}{5 + 4 \sin x} dx = A \tan^{- 1} \left( B \tan\frac{x}{2} + \frac{4}{3} \right) + C,\] then
\[\int\frac{1}{1 + \tan x} dx =\]
\[\int\frac{\sin x}{3 + 4 \cos^2 x} dx\]
\[\int\frac{\cos2x - \cos2\theta}{\cos x - \cos\theta}dx\] is equal to
\[\int\frac{x^3}{x + 1}dx\] is equal to
\[\int\frac{\sin x}{1 + \sin x} \text{ dx }\]
\[\int\sqrt{\frac{1 - x}{x}} \text{ dx}\]
\[\int \tan^3 x\ \sec^4 x\ dx\]
\[\int\frac{1}{x\sqrt{1 + x^3}} \text{ dx}\]
\[\int \sec^{- 1} \sqrt{x}\ dx\]
