Advertisements
Advertisements
प्रश्न
Evaluate:\[\int\frac{x^2}{1 + x^3} \text{ dx }\] .
Advertisements
उत्तर
\[\text{ Let I }= \int \frac{x^2 dx}{1 + x^3}\]
\[\text{ Putting 1} + x^3 = t\]
\[ \Rightarrow 3 x^2 \text{ dx} = dt\]
\[ \Rightarrow x^2 \text{ dx} = \frac{dt}{3}\]
\[ \therefore I = \frac{1}{3}\int \frac{dt}{t}\]
\[ = \frac{1}{3}\text{ ln } \left| t \right| + C\]
\[ = \frac{1}{3}\text{ ln} \left| 1 + x^3 \right| + C \left( \because t = 1 + x^3 \right)\]
APPEARS IN
संबंधित प्रश्न
Evaluate the following integrals:
` ∫ cot^3 x "cosec"^2 x dx `
Evaluate the following integrals:
Evaluate the following integrals:
Evaluate the following integrals:
Evaluate the following integrals:
Evaluate the following integral:
Evaluate the following integrals:
Evaluate the following integral:
Evaluate:\[\int\frac{\left( 1 + \log x \right)^2}{x} \text{ dx }\]
Evaluate:\[\int\frac{e\tan^{- 1} x}{1 + x^2} \text{ dx }\]
Evaluate: \[\int\frac{1}{\sqrt{1 - x^2}} \text{ dx }\]
Evaluate: \[\int\frac{1}{x^2 + 16}\text{ dx }\]
Evaluate: `int_ (x + sin x)/(1 + cos x ) dx`
Evaluate the following:
`int ("d"x)/sqrt(16 - 9x^2)`
Evaluate the following:
`int (3x - 1)/sqrt(x^2 + 9) "d"x`
Evaluate the following:
`int sqrt(2"a"x - x^2) "d"x`
Evaluate the following:
`int sqrt(x)/(sqrt("a"^3 - x^3)) "d"x`
