Advertisements
Advertisements
प्रश्न
The derivative of \[f\left( x \right) = \int\limits_{x^2}^{x^3} \frac{1}{\log_e t} dt, \left( x > 0 \right),\] is
विकल्प
- \[\frac{1}{3 \ln x}\]
- \[\frac{1}{3 \ln x} - \frac{1}{2 \ln x}\]
(ln x)−1 x (x − 1)
- \[\frac{3 x^2}{\ln x}\]
Advertisements
उत्तर
(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
संबंधित प्रश्न
Evaluate the following definite integrals:
If `f` is an integrable function such that f(2a − x) = f(x), then prove that
Prove that:
\[\int\limits_0^{2a} f\left( x \right) dx\] is equal to
The value of \[\int\limits_0^1 \tan^{- 1} \left( \frac{2x - 1}{1 + x - x^2} \right) dx,\] is
Evaluate : \[\int\limits_0^{2\pi} \cos^5 x dx\] .
Evaluate: \[\int\limits_{- \pi/2}^{\pi/2} \frac{\cos x}{1 + e^x}dx\] .
\[\int\limits_1^2 x\sqrt{3x - 2} dx\]
\[\int\limits_0^{\pi/3} \frac{\cos x}{3 + 4 \sin x} dx\]
\[\int\limits_{- 1/2}^{1/2} \cos x \log\left( \frac{1 + x}{1 - x} \right) dx\]
\[\int\limits_{- \pi/4}^{\pi/4} \left| \tan x \right| dx\]
\[\int\limits_0^\pi \frac{x}{a^2 - \cos^2 x} dx, a > 1\]
\[\int\limits_1^3 \left( x^2 + 3x \right) dx\]
Evaluate the following:
`int_(-1)^1 "f"(x) "d"x` where f(x) = `{{:(x",", x ≥ 0),(-x",", x < 0):}`
Evaluate the following integrals as the limit of the sum:
`int_0^1 (x + 4) "d"x`
Choose the correct alternative:
If n > 0, then Γ(n) is
Evaluate `int (x^2"d"x)/(x^4 + x^2 - 2)`
Verify the following:
`int (2x + 3)/(x^2 + 3x) "d"x = log|x^2 + 3x| + "C"`
