Advertisements
Advertisements
Question
`int "e"^x ((1 - x)/(1 + x^2))^2 "d"x` is equal to ______.
Options
`"e"^x/(1 + x^2) + "C"`
`(-"e"^x)/(1 + x^2) + "C"`
`"e"^x/(1 + x^2)^2 + "C"`
`(-"e"^x)/(1 + x^2)^2 + "C"`
Advertisements
Solution
`int "e"^x ((1 - x)/(1 + x^2))^2 "d"x` is equal to `"e"^x/(1 + x^2) + "C"`.
Explanation:
Let I = `int "e"^x ((1 - x)/(1 + x^2))^2 "d"x`
= `int "e"^x [(1 + x^2 - 2x)/(1 + x^2)^2]"d"x`
= `int "e"^x [((1 + x^2))/(1 + x^2)^2 - (2x)/(1 + x^2)^2]"d"x`
= `int "e"^x [1/(1 + x^2) - (2x)/(1 + x^2)^2]"d"x`
Here f(x) = `1/(1 + x^2)`
∴ f'(x) = `(-2x)/(1 + x^2)^2`
Using `int "e"^x ["f"(x) + "f'"(x)]"d"x = "e"^x * "f"(x) + "C"`
∴ I = `"e"^x * 1/(1 + x^2) + "C" = "e"^x/(1 + x^2) + "C"`
APPEARS IN
RELATED QUESTIONS
Evaluate each of the following integral:
If \[\int_0^a \frac{1}{4 + x^2}dx = \frac{\pi}{8}\] , find the value of a.
The value of \[\int\limits_0^\pi \frac{x \tan x}{\sec x + \cos x} dx\] is __________ .
\[\int\limits_0^{\pi/2} \frac{1}{2 + \cos x} dx\] equals
`int_0^1 sqrt((1 - "x")/(1 + "x")) "dx"`
The value of \[\int\limits_0^\pi \frac{1}{5 + 3 \cos x} dx\] is
If f (a + b − x) = f (x), then \[\int\limits_a^b\] x f (x) dx is equal to
Evaluate : \[\int\limits_0^\pi/4 \frac{\sin x + \cos x}{16 + 9 \sin 2x}dx\] .
`int_0^(2a)f(x)dx`
\[\int\limits_1^2 \frac{x + 3}{x\left( x + 2 \right)} dx\]
\[\int\limits_0^\pi \frac{x}{a^2 \cos^2 x + b^2 \sin^2 x} dx\]
\[\int\limits_0^{\pi/2} \frac{x \sin x \cos x}{\sin^4 x + \cos^4 x} dx\]
\[\int\limits_2^3 \frac{\sqrt{x}}{\sqrt{5 - x} + \sqrt{x}} dx\]
\[\int\limits_{- \pi}^\pi x^{10} \sin^7 x dx\]
Evaluate the following:
`int_1^4` f(x) dx where f(x) = `{{:(4x + 3",", 1 ≤ x ≤ 2),(3x + 5",", 2 < x ≤ 4):}`
Evaluate the following using properties of definite integral:
`int_0^(i/2) (sin^7x)/(sin^7x + cos^7x) "d"x`
Evaluate the following:
`int ((x^2 + 2))/(x + 1) "d"x`
Find: `int logx/(1 + log x)^2 dx`
