Advertisements
Advertisements
प्रश्न
पर्याय
\[ x + \frac{x^2}{2} + \frac{x^3}{3} - \log\left| 1 - x \right| + C\]
\[ x + \frac{x^2}{2} - \frac{x^3}{3} - \log\left| 1 - x \right| + C\]
\[ x - \frac{x^2}{2} - \frac{x^3}{3} - \log\left| 1 + x \right| + C\]
- \[ x - \frac{x^2}{2} + \frac{x^3}{3} - \log\left| 1 + x \right| + C\]
उत्तर
\[ x - \frac{x^2}{2} + \frac{x^3}{3} - \log\left| 1 + x \right| + C\]
\[\text{Let }I = \int\frac{x^3}{x + 1}dx\]
\[ = \int\frac{x^3 + 1 - 1}{x + 1}dx\]
\[ = \int\left( \frac{x^3 + 1}{x + 1} - \frac{1}{x + 1} \right)dx\]
\[ = \int\left( \frac{\left( x + 1 \right)\left( x^2 - x + 1 \right)}{x + 1} - \frac{1}{x + 1} \right)dx\]
\[ = \int\left( x^2 - x + 1 - \frac{1}{x + 1} \right)dx\]
\[ = \left( \frac{x^3}{3} - \frac{x^2}{2} + x - \log\left| x + 1 \right| \right) + C\]
\[ = \frac{x^3}{3} - \frac{x^2}{2} + x - \log\left| x + 1 \right| + C\]
\[\text{Therefore, }\int\frac{x^3}{x + 1}dx = \frac{x^3}{3} - \frac{x^2}{2} + x - \log\left| x + 1 \right| + C\]
APPEARS IN
संबंधित प्रश्न
` ∫ tan x sec^4 x dx `
` ∫ \sqrt{tan x} sec^4 x dx `
Evaluate the following integral:
\[\int\frac{1}{4 \sin^2 x + 4 \sin x \cos x + 5 \cos^2 x} \text{ dx }\]
\[ \int\left( 1 + x^2 \right) \ \cos 2x \ dx\]
