Advertisements
Advertisements
प्रश्न
Write a value of
\[\int e^{3 \text{ log x}} x^4\text{ dx}\]
योग
Advertisements
उत्तर
\[\int e^{3 \text{ log x}} . x^4 \text{ dx }\]
\[ = \int e^{\text{ log x}^3} \cdot \text{ x}^4\text{ dx } \left( \because a\log x = \log x^a \right)\]
\[ = \int x^3 \cdot x^4 \text{ dx } \left( \because e^{\text{ log m}} = m \right)\]
\[ = \int x^7 \cdot dx\]
\[ = \frac{x^8}{8} + C\]
shaalaa.com
क्या इस प्रश्न या उत्तर में कोई त्रुटि है?
APPEARS IN
संबंधित प्रश्न
\[\int\left\{ x^2 + e^{\log x}+ \left( \frac{e}{2} \right)^x \right\} dx\]
\[\int\frac{\left( 1 + \sqrt{x} \right)^2}{\sqrt{x}} dx\]
\[\int\frac{\tan x}{\sec x + \tan x} dx\]
\[\int \left( a \tan x + b \cot x \right)^2 dx\]
\[\int\frac{1 - \cos x}{1 + \cos x} dx\]
`∫ cos ^4 2x dx `
\[\int\sqrt{\frac{1 + \cos 2x}{1 - \cos 2x}} dx\]
\[\int\frac{\text{sin} \left( x - \alpha \right)}{\text{sin }\left( x + \alpha \right)} dx\]
\[\int\frac{1 - \cot x}{1 + \cot x} dx\]
\[\int\frac{\cos x - \sin x}{1 + \sin 2x} dx\]
\[\int\frac{\left( x + 1 \right) e^x}{\sin^2 \left( \text{x e}^x \right)} dx\]
\[\int\frac{\text{sin }\left( \text{2 + 3 log x }\right)}{x} dx\]
` ∫ tan x sec^4 x dx `
\[\int\frac{x}{x^4 + 2 x^2 + 3} dx\]
\[\int\frac{x^2}{x^6 + a^6} dx\]
\[\int\frac{1}{\sqrt{5 - 4x - 2 x^2}} dx\]
\[\int\frac{\left( 3\sin x - 2 \right)\cos x}{13 - \cos^2 x - 7\sin x}dx\]
\[\int\frac{1}{4 \cos x - 1} \text{ dx }\]
\[\int\frac{1}{p + q \tan x} \text{ dx }\]
\[\int\frac{2 \tan x + 3}{3 \tan x + 4} \text{ dx }\]
\[\int \tan^{- 1} \left( \sqrt{x} \right) \text{dx }\]
\[\int \sin^3 \sqrt{x}\ dx\]
\[\int e^x \left( \frac{1}{x^2} - \frac{2}{x^3} \right) dx\]
\[\int e^x \left( \frac{\sin 4x - 4}{1 - \cos 4x} \right) dx\]
\[\int\left( x + 1 \right) \sqrt{2 x^2 + 3} \text{ dx}\]
\[\int x\sqrt{x^2 + x} \text{ dx }\]
\[\int\frac{3}{\left( 1 - x \right) \left( 1 + x^2 \right)} dx\]
\[\int\frac{x + 1}{x \left( 1 + x e^x \right)} dx\]
\[\int\frac{1}{\cos x + \sqrt{3} \sin x} \text{ dx } \] is equal to
\[\int\frac{1}{7 + 5 \cos x} dx =\]
The value of \[\int\frac{\sin x + \cos x}{\sqrt{1 - \sin 2x}} dx\] is equal to
\[\int\frac{\cos2x - \cos2\theta}{\cos x - \cos\theta}dx\] is equal to
\[\int\frac{x^3}{x + 1}dx\] is equal to
\[\int\frac{\sin x}{1 + \sin x} \text{ dx }\]
\[\int\sin x \sin 2x \text{ sin 3x dx }\]
\[\int\text{ cos x cos 2x cos 3x dx}\]
\[\int\frac{5x + 7}{\sqrt{\left( x - 5 \right) \left( x - 4 \right)}} \text{ dx }\]
\[\int\frac{\sin^6 x}{\cos x} \text{ dx }\]
\[\int\frac{1}{\left( x^2 + 2 \right) \left( x^2 + 5 \right)} \text{ dx}\]
