Advertisements
Advertisements
Question
Advertisements
Solution
\[\text{Let I} = \int\frac{10 x^9 + {10}^x \log_e 10}{{10}^x + x^{10}}dx\]
\[\text{Putting }{10}^x + x^{10} = t\]
\[ \Rightarrow {10}^x \log_e 10 + 10 x^9 = \frac{dt}{dx}\]
\[ \Rightarrow \left( {10}^x \log_e 10 + 10 x^9 \right)dx = dt\]
\[ \therefore I = \int\frac{1}{t}dt\]
\[ = \text{ln} \left| t \right| + C\]
\[ = \text{ln} \left| {10}^x + x^{10} \right| + C \left[ \because t = {10}^x + x^{10} \right]\]
APPEARS IN
RELATED QUESTIONS
` ∫ cot^3 x "cosec"^2 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 integral:
Evaluate the following integral:
Write a value of
Evaluate:\[\int\frac{\sin \sqrt{x}}{\sqrt{x}} \text{ dx }\]
Evaluate:\[\int\frac{\cos \sqrt{x}}{\sqrt{x}} \text{ dx }\]
Evaluate:\[\int\frac{\left( 1 + \log x \right)^2}{x} \text{ dx }\]
Evaluate:\[\int \sec^2 \left( 7 - 4x \right) \text{ dx }\]
Evaluate:\[\int\frac{\log x}{x} \text{ dx }\]
Evaluate: \[\int 2^x \text{ dx }\]
Evaluate: \[\int\frac{x^3 - x^2 + x - 1}{x - 1} \text{ dx }\]
Write the value of\[\int\sec x \left( \sec x + \tan x \right)\text{ dx }\]
Evaluate: \[\int\frac{x + \cos6x}{3 x^2 + \sin6x}\text{ dx }\]
Evaluate : \[\int\frac{1}{x(1 + \log x)} \text{ dx}\]
Evaluate the following:
`int ("d"x)/sqrt(16 - 9x^2)`
Evaluate the following:
`int sqrt(5 - 2x + x^2) "d"x`
Evaluate the following:
`int sqrt(2"a"x - x^2) "d"x`
Evaluate the following:
`int_1^2 ("d"x)/sqrt((x - 1)(2 - x))`
