Advertisements
Advertisements
प्रश्न
Evaluate:\[\int\frac{\left( 1 + \log x \right)^2}{x} \text{ dx }\]
Advertisements
उत्तर
\[\text{ Let } \int\frac{\left( 1 + \log x \right)^2}{x} \text{ dx }\]
\[\text{ Putting 1} + \log x = t\]
\[ \Rightarrow \frac{1}{x} dx = dt\]
\[ \therefore I = \int t^2 \cdot dt\]
\[ = \frac{t^3}{3} + C\]
\[ = \frac{\left( 1 + \log x \right)^3}{3} + C \left[ \because t = \left( 1 + \log x \right) \right]\]
APPEARS IN
संबंधित प्रश्न
Evaluate the following integrals:
` ∫ cot^3 x "cosec"^2 x dx `
\[\int\frac{\left\{ e^{\sin^{- 1} }x \right\}^2}{\sqrt{1 - x^2}} dx\]
Evaluate the following integrals:
Evaluate the following integral:
Evaluate the following integral:
Evaluate the following integral:
Evaluate:\[\int\frac{\cos \sqrt{x}}{\sqrt{x}} \text{ dx }\]
Evaluate:\[\int\frac{\log x}{x} \text{ dx }\]
Evaluate: \[\int 2^x \text{ dx }\]
Write the value of\[\int\sec x \left( \sec x + \tan x \right)\text{ dx }\]
Evaluate: \[\int\left( 1 - x \right)\sqrt{x}\text{ dx }\]
Evaluate: \[\int\frac{x + \cos6x}{3 x^2 + \sin6x}\text{ dx }\]
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`
Evaluate the following:
`int ("d"x)/(xsqrt(x^4 - 1))` (Hint: Put x2 = sec θ)
Evaluate the following:
`int_1^2 ("d"x)/sqrt((x - 1)(2 - x))`
