Advertisements
Advertisements
Question
The derivative of \[f\left( x \right) = \int\limits_{x^2}^{x^3} \frac{1}{\log_e t} dt, \left( x > 0 \right),\] is
Options
- \[\frac{1}{3 \ln x}\]
- \[\frac{1}{3 \ln x} - \frac{1}{2 \ln x}\]
(ln x)−1 x (x − 1)
- \[\frac{3 x^2}{\ln x}\]
Advertisements
Solution
(ln x)−1 x (x − 1)
Using Newton Leibnitz formula
\[f' (x) = \frac{1}{\log_e x^3}(3 x^2 ) - \frac{1}{\log_e x^2}(2x) \]
\[= \frac{3 x^2}{3\ln x}- \frac{2x}{2\ln x} \]
\[= \frac{x^2}{\ln x} - \frac{x}{\ln x} \]
\[= \frac{1}{\ln x}x(x - 1) \]
\[= {(\ln x)}^{- 1} x(x - 1)\]
APPEARS IN
RELATED QUESTIONS
\[\int\limits_0^\pi \frac{1}{1 + \sin x} dx\] equals
The value of the integral \[\int\limits_0^\infty \frac{x}{\left( 1 + x \right)\left( 1 + x^2 \right)} dx\]
\[\int\limits_0^{2a} f\left( x \right) dx\] is equal to
\[\int\limits_0^1 \tan^{- 1} x dx\]
\[\int\limits_0^{\pi/2} x^2 \cos 2x dx\]
\[\int\limits_0^{15} \left[ x^2 \right] dx\]
\[\int\limits_{- 1}^1 e^{2x} dx\]
Evaluate the following using properties of definite integral:
`int_0^1 log (1/x - 1) "d"x`
Evaluate the following integrals as the limit of the sum:
`int_1^3 x "d"x`
Evaluate `int (x^2"d"x)/(x^4 + x^2 - 2)`
Given `int "e"^"x" (("x" - 1)/("x"^2)) "dx" = "e"^"x" "f"("x") + "c"`. Then f(x) satisfying the equation is:
What is the meaning of an indefinite integral?
