Advertisements
Advertisements
प्रश्न
Advertisements
उत्तर
\[Let\ I = \int_0^1 \frac{1 - x^2}{\left( 1 + x^2 \right)^2} d\ x . Then, \]
\[I = \int_0^1 \frac{\left( \frac{1}{x^2} - 1 \right)}{\left( x + \frac{1}{x} \right)^2} dx\]
\[Let\ x + \frac{1}{x} = t . Then, 1 - \frac{1}{x^2} dx = dt\]
\[When\ x = 0, t\ = \infty\ and\ x\ = 1, t = 2\]
\[ \therefore I = \int_\infty^2 \frac{- dt}{t^2}\]
\[ \Rightarrow I = \left[ \frac{1}{t} \right]_\infty^2 \]
\[ \Rightarrow I = \frac{1}{2} - 0\]
\[ \Rightarrow I = \frac{1}{2}\]
APPEARS IN
संबंधित प्रश्न
If \[\int\limits_0^a \frac{1}{1 + 4 x^2} dx = \frac{\pi}{8},\] then a equals
Evaluate : \[\int\frac{dx}{\sin^2 x \cos^2 x}\] .
\[\int\limits_0^{\pi/2} \left| \sin x - \cos x \right| dx\]
\[\int\limits_{- \pi/2}^{\pi/2} \sin^9 x dx\]
Using second fundamental theorem, evaluate the following:
`int_0^1 x"e"^(x^2) "d"x`
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):}`
Evaluate the following using properties of definite integral:
`int_(- pi/2)^(pi/2) sin^2theta "d"theta`
Choose the correct alternative:
Γ(n) is
Choose the correct alternative:
If n > 0, then Γ(n) is
Given `int "e"^"x" (("x" - 1)/("x"^2)) "dx" = "e"^"x" "f"("x") + "c"`. Then f(x) satisfying the equation is:
What is the meaning of an indefinite integral?
