Advertisements
Advertisements
प्रश्न
Find: `int logx/(1 + log x)^2 dx`
Advertisements
उत्तर
`int logx/(1 + log x)^2 dx = int (log x + 1 - 1)/(1 + log x)^2 dx`
= `int 1/(1 + log x) dx - int 1/(1 + log x)^2 dx`
= `1/(1 + log x) xx x - int (-1)/(1 + log x)^2 xx 1/x xx xdx - int 1/(1 + log x)^2 dx`
= ` x/(1 + log x) + c`
APPEARS IN
संबंधित प्रश्न
Evaluate the following definite integrals:
Evaluate each of the following integral:
\[\int_a^b \frac{x^\frac{1}{n}}{x^\frac{1}{n} + \left( a + b - x \right)^\frac{1}{n}}dx, n \in N, n \geq 2\]
If \[f\left( x \right) = \int_0^x t\sin tdt\], the write the value of \[f'\left( x \right)\]
The value of the integral \[\int\limits_0^\infty \frac{x}{\left( 1 + x \right)\left( 1 + x^2 \right)} dx\]
\[\int\limits_0^4 x\sqrt{4 - x} dx\]
\[\int\limits_0^{\pi/2} \frac{\sin x}{\sqrt{1 + \cos x}} dx\]
\[\int\limits_{- 1/2}^{1/2} \cos x \log\left( \frac{1 + x}{1 - x} \right) dx\]
\[\int\limits_0^{\pi/2} \frac{1}{1 + \tan^3 x} dx\]
\[\int\limits_0^\pi x \sin x \cos^4 x dx\]
\[\int\limits_0^{\pi/2} \frac{\cos^2 x}{\sin x + \cos x} dx\]
Evaluate the following:
`int_0^oo "e"^(-mx) x^6 "d"x`
Evaluate the following integrals as the limit of the sum:
`int_1^3 (2x + 3) "d"x`
Choose the correct alternative:
`int_0^1 (2x + 1) "d"x` is
Choose the correct alternative:
Γ(1) is
