Advertisements
Advertisements
प्रश्न
The value of `int_2^3 x/(x^2 + 1)`dx is ______.
विकल्प
`log 4`
`log 3/2`
`1/2 log2`
`log 9/4`
Advertisements
उत्तर
The value of `int_2^3 x/(x^2 + 1)`dx is `underline(bb(1/2 log 2))`.
Explanation:
`int_2^3 x/(x^2 + 1) = 1/2 [log(x^2 + 1)]_2^3`
= `1/2 (log 10 - log 5)`
= `1/2 log (10/5)`
= `1/2 log 2`
APPEARS IN
संबंधित प्रश्न
If \[f\left( a + b - x \right) = f\left( x \right)\] , then prove that \[\int_a^b xf\left( x \right)dx = \frac{a + b}{2} \int_a^b f\left( x \right)dx\]
Solve each of the following integral:
\[\int\limits_0^\pi \frac{1}{1 + \sin x} dx\] equals
The value of \[\int\limits_0^\pi \frac{1}{5 + 3 \cos x} dx\] is
\[\int\limits_0^4 x\sqrt{4 - x} dx\]
\[\int\limits_0^1 \cos^{- 1} x dx\]
\[\int\limits_0^\infty \frac{x}{\left( 1 + x \right)\left( 1 + x^2 \right)} dx\]
\[\int\limits_{- 1/2}^{1/2} \cos x \log\left( \frac{1 + x}{1 - x} \right) dx\]
\[\int\limits_0^a \frac{\sqrt{x}}{\sqrt{x} + \sqrt{a - x}} dx\]
\[\int\limits_0^{\pi/2} \frac{\cos^2 x}{\sin x + \cos x} dx\]
\[\int\limits_0^\pi \cos 2x \log \sin x dx\]
\[\int\limits_0^\pi \frac{x}{a^2 - \cos^2 x} dx, a > 1\]
\[\int\limits_0^1 \cot^{- 1} \left( 1 - x + x^2 \right) dx\]
\[\int\limits_{\pi/6}^{\pi/2} \frac{\ cosec x \cot x}{1 + {cosec}^2 x} dx\]
\[\int\limits_0^4 x dx\]
Choose the correct alternative:
The value of `int_(- pi/2)^(pi/2) cos x "d"x` is
Find `int sqrt(10 - 4x + 4x^2) "d"x`
