Advertisements
Advertisements
प्रश्न
Solve each of the following integral:
Advertisements
उत्तर
\[\int_2^4 \frac{x}{x^2 + 1}dx\]
\[ = \frac{1}{2} \int_2^4 \frac{2x}{x^2 + 1}dx\]
\[ = \frac{1}{2} \times \left.\log\left( x^2 + 1 \right)\right|_2^4 ...................\left[ \int\frac{f'\left( x \right)}{f\left( x \right)}dx = \log f\left( x \right) + C \right]\]
\[ = \frac{1}{2}\left( \log17 - \log5 \right)\]
\[ = \frac{1}{2}\log\left( \frac{17}{5} \right) .............\left( \log a - \log b = \log\frac{a}{b} \right)\]
APPEARS IN
संबंधित प्रश्न
If \[f\left( a + b - x \right) = f\left( x \right)\] , then prove that \[\int_a^b xf\left( x \right)dx = \frac{a + b}{2} \int_a^b f\left( x \right)dx\]
Evaluate each of the following integral:
\[\int\limits_0^1 \frac{1 - x}{1 + x} dx\]
\[\int\limits_0^1 \sqrt{\frac{1 - x}{1 + x}} dx\]
\[\int\limits_1^2 \frac{x + 3}{x\left( x + 2 \right)} dx\]
\[\int\limits_1^3 \left| x^2 - 2x \right| dx\]
\[\int\limits_{- \pi/2}^{\pi/2} \sin^9 x dx\]
\[\int\limits_0^\pi \frac{x \tan x}{\sec x + \tan x} dx\]
Find : `∫_a^b logx/x` dx
Choose the correct alternative:
`int_0^oo x^4"e"^-x "d"x` is
If x = `int_0^y "dt"/sqrt(1 + 9"t"^2)` and `("d"^2y)/("d"x^2)` = ay, then a equal to ______.
`int "e"^x ((1 - x)/(1 + x^2))^2 "d"x` is equal to ______.
`int x^9/(4x^2 + 1)^6 "d"x` is equal to ______.
