Advertisements
Advertisements
Question
\[\int\limits_{- a}^a \frac{x e^{x^2}}{1 + x^2} dx\]
Advertisements
Solution
\[\int_{- a}^a \frac{x e^{x^2}}{1 + x^2} d x\]
\[\text{Let }f(x) = \frac{x e^{x^2}}{1 + x^2}\]
\[\text{Consider }f(-x) = - \frac{x e^{x^2}}{1 + x^2} = - f\left( x \right)\]
Thus f(x) is an odd function
Therefore,
\[ \int_{- a}^a \frac{x e^{x^2}}{1 + x^2} d x = 0\]
APPEARS IN
RELATED QUESTIONS
Evaluate each of the following integral:
\[\int_a^b \frac{x^\frac{1}{n}}{x^\frac{1}{n} + \left( a + b - x \right)^\frac{1}{n}}dx, n \in N, n \geq 2\]
Solve each of the following integral:
The value of \[\int\limits_0^\pi \frac{x \tan x}{\sec x + \cos x} dx\] is __________ .
Given that \[\int\limits_0^\infty \frac{x^2}{\left( x^2 + a^2 \right)\left( x^2 + b^2 \right)\left( x^2 + c^2 \right)} dx = \frac{\pi}{2\left( a + b \right)\left( b + c \right)\left( c + a \right)},\] the value of \[\int\limits_0^\infty \frac{dx}{\left( x^2 + 4 \right)\left( x^2 + 9 \right)},\]
If \[\int\limits_0^a \frac{1}{1 + 4 x^2} dx = \frac{\pi}{8},\] then a equals
If \[\int\limits_0^1 f\left( x \right) dx = 1, \int\limits_0^1 xf\left( x \right) dx = a, \int\limits_0^1 x^2 f\left( x \right) dx = a^2 , then \int\limits_0^1 \left( a - x \right)^2 f\left( x \right) dx\] equals
The derivative of \[f\left( x \right) = \int\limits_{x^2}^{x^3} \frac{1}{\log_e t} dt, \left( x > 0 \right),\] is
Evaluate : \[\int\limits_0^\pi \frac{x}{1 + \sin \alpha \sin x}dx\] .
\[\int\limits_1^5 \frac{x}{\sqrt{2x - 1}} dx\]
\[\int\limits_0^{\pi/2} \frac{\cos x}{1 + \sin^2 x} dx\]
\[\int\limits_0^a \frac{\sqrt{x}}{\sqrt{x} + \sqrt{a - x}} dx\]
\[\int\limits_{- \pi}^\pi x^{10} \sin^7 x dx\]
\[\int\limits_2^3 e^{- x} dx\]
Evaluate the following using properties of definite integral:
`int_0^1 x/((1 - x)^(3/4)) "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^oo "e"^(-2x) "d"x` is
Choose the correct alternative:
`int_(-1)^1 x^3 "e"^(x^4) "d"x` is
Choose the correct alternative:
If n > 0, then Γ(n) is
Evaluate `int "dx"/sqrt((x - alpha)(beta - x)), beta > alpha`
Given `int "e"^"x" (("x" - 1)/("x"^2)) "dx" = "e"^"x" "f"("x") + "c"`. Then f(x) satisfying the equation is:
Find: `int logx/(1 + log x)^2 dx`
The value of `int_2^3 x/(x^2 + 1)`dx is ______.
