Advertisements
Advertisements
प्रश्न
`int "e"^x ((1 - x)/(1 + x^2))^2 "d"x` is equal to ______.
विकल्प
`"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
उत्तर
`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
संबंधित प्रश्न
Evaluate each of the following integral:
Evaluate :
The value of \[\int\limits_0^\pi \frac{x \tan x}{\sec x + \cos x} dx\] is __________ .
The value of the integral \[\int\limits_0^\infty \frac{x}{\left( 1 + x \right)\left( 1 + x^2 \right)} dx\]
If \[\int\limits_0^a \frac{1}{1 + 4 x^2} dx = \frac{\pi}{8},\] then a equals
Evaluate : \[\int\limits_0^\pi/4 \frac{\sin x + \cos x}{16 + 9 \sin 2x}dx\] .
\[\int\limits_0^1 \tan^{- 1} x dx\]
\[\int\limits_0^1 \cos^{- 1} \left( \frac{1 - x^2}{1 + x^2} \right) dx\]
\[\int\limits_2^3 \frac{\sqrt{x}}{\sqrt{5 - x} + \sqrt{x}} dx\]
Evaluate the following integrals as the limit of the sum:
`int_0^1 x^2 "d"x`
If x = `int_0^y "dt"/sqrt(1 + 9"t"^2)` and `("d"^2y)/("d"x^2)` = ay, then a equal to ______.
Which notation represents a definite integral?
