Advertisements
Advertisements
प्रश्न
उत्तर
\[\text{Let I} = \int\frac{10 x^9 + {10}^x \log_e 10}{{10}^x + x^{10}}dx\]
\[\text{Putting }{10}^x + x^{10} = t\]
\[ \Rightarrow {10}^x \log_e 10 + 10 x^9 = \frac{dt}{dx}\]
\[ \Rightarrow \left( {10}^x \log_e 10 + 10 x^9 \right)dx = dt\]
\[ \therefore I = \int\frac{1}{t}dt\]
\[ = \text{ln} \left| t \right| + C\]
\[ = \text{ln} \left| {10}^x + x^{10} \right| + C \left[ \because t = {10}^x + x^{10} \right]\]
APPEARS IN
संबंधित प्रश्न
Evaluate the following integrals:
\[\int\frac{\left\{ e^{\sin^{- 1} }x \right\}^2}{\sqrt{1 - x^2}} dx\]
\[\int\frac{1}{\sqrt{1 - x^2} \left( \sin^{- 1} x \right)^2} dx\]
\[\int\frac{\cot x}{\sqrt{\sin x}} dx\]
Evaluate the following integrals:
Evaluate the following integrals:
Evaluate the following integrals:
Evaluate the following integral :-
Evaluate the following integral:
Evaluate the following integral:
Evaluate the following integrals:
Evaluate the following integral:
Evaluate the following integral:
Evaluate:
Evaluate:\[\int\frac{\sec^2 \sqrt{x}}{\sqrt{x}} \text{ dx }\]
Evaluate:\[\int\frac{\cos \sqrt{x}}{\sqrt{x}} \text{ dx }\]
Evaluate:\[\int\frac{\left( 1 + \log x \right)^2}{x} \text{ dx }\]
Evaluate:\[\int\frac{\log x}{x} \text{ dx }\]
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: \[\int\frac{2}{1 - \cos2x}\text{ dx }\]
Evaluate:
Evaluate the following:
`int sqrt(1 + x^2)/x^4 "d"x`
Evaluate the following:
`int ("d"x)/sqrt(16 - 9x^2)`
Evaluate the following:
`int sqrt(2"a"x - x^2) "d"x`
Evaluate the following:
`int ("d"x)/(xsqrt(x^4 - 1))` (Hint: Put x2 = sec θ)