Advertisements
Advertisements
प्रश्न
उत्तर
\[\int\left( \frac{x + 1}{x} \right) \cdot \left( x + \log x \right)^2 dx\]
\[\text{Let x} + \log x = t\]
\[ \Rightarrow \left( 1 + \frac{1}{x} \right) = \frac{dt}{dx}\]
\[ \Rightarrow \left( \frac{x + 1}{x} \right) dx = dt\]
\[Now, \int\left( \frac{x + 1}{x} \right) \cdot \left( x + \log x \right)^2 dx\]
\[ = \int t^2 dt\]
\[ = \frac{t^3}{3} + C\]
\[ = \frac{\left( x + \log x \right)^3}{3} + C\]
APPEARS IN
संबंधित प्रश्न
`∫ cos ^4 2x dx `
` ∫ tan^5 x dx `
` ∫ {x-3} /{ x^2 + 2x - 4 } dx `
\[\int\left( e^\text{log x} + \sin x \right) \text{ cos x dx }\]
The value of \[\int\frac{\sin x + \cos x}{\sqrt{1 - \sin 2x}} dx\] is equal to
\[\int\frac{1}{\sqrt{x} + \sqrt{x + 1}} \text{ dx }\]
\[\int\frac{1}{\sin^4 x + \cos^4 x} \text{ dx}\]
\[\int\frac{1}{2 + \cos x} \text{ dx }\]
\[\int\frac{5 x^4 + 12 x^3 + 7 x^2}{x^2 + x} dx\]