Advertisements
Advertisements
प्रश्न
उत्तर
\[Let\ x^2 = t . Then, 2x\ dx = dt\]
\[When\ x = 1, t = 1\ and\ x = 2, t = 4\]
\[ \therefore I = \int_1^2 \frac{3x}{9 x^2 - 1} d x\]
\[ \Rightarrow I = \frac{3}{2} \int_1^4 \frac{dt}{9t - 1}\]
\[ \Rightarrow I = \frac{3}{18} \left[ \log \left( 9t - 1 \right) \right]_1^4 \]
\[ \Rightarrow I = \frac{3}{18}\left( \log 35 - \log 8 \right)\]
\[ \Rightarrow I = \frac{\left( \log 35 - \log 8 \right)}{6}\]
APPEARS IN
संबंधित प्रश्न
\[\int\limits_{\pi/4}^{\pi/2} \cot x\ dx\]
Evaluate the following integral:
If f is an integrable function, show that
If \[\int\limits_0^a 3 x^2 dx = 8,\] write the value of a.
\[\int\limits_1^2 x\sqrt{3x - 2} dx\]
\[\int\limits_0^1 \cos^{- 1} \left( \frac{1 - x^2}{1 + x^2} \right) dx\]
\[\int\limits_0^1 \frac{1 - x}{1 + x} dx\]
\[\int\limits_0^{\pi/4} \sin 2x \sin 3x dx\]
\[\int\limits_0^{\pi/2} \left| \sin x - \cos x \right| dx\]
\[\int\limits_0^\pi \frac{x \sin x}{1 + \cos^2 x} dx\]
\[\int\limits_0^{\pi/2} \frac{x \sin x \cos x}{\sin^4 x + \cos^4 x} dx\]
\[\int\limits_{- \pi}^\pi x^{10} \sin^7 x dx\]
\[\int\limits_0^2 \left( 2 x^2 + 3 \right) dx\]
Evaluate the following:
`int_0^2 "f"(x) "d"x` where f(x) = `{{:(3 - 2x - x^2",", x ≤ 1),(x^2 + 2x - 3",", 1 < x ≤ 2):}`
Choose the correct alternative:
`int_0^oo x^4"e"^-x "d"x` is
`int (x + 3)/(x + 4)^2 "e"^x "d"x` = ______.
