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 the following integral:
If f(2a − x) = −f(x), prove that
If f is an integrable function, show that
Evaluate each of the following integral:
The value of \[\int\limits_{- \pi/2}^{\pi/2} \left( x^3 + x \cos x + \tan^5 x + 1 \right) dx, \] is
\[\int\limits_0^1 \cos^{- 1} \left( \frac{1 - x^2}{1 + x^2} \right) dx\]
\[\int\limits_0^\pi \sin^3 x\left( 1 + 2 \cos x \right) \left( 1 + \cos x \right)^2 dx\]
\[\int\limits_0^\infty \frac{x}{\left( 1 + x \right)\left( 1 + x^2 \right)} dx\]
\[\int\limits_1^3 \left| x^2 - 2x \right| dx\]
\[\int\limits_0^1 \cot^{- 1} \left( 1 - x + x^2 \right) dx\]
\[\int\limits_{\pi/6}^{\pi/2} \frac{\ cosec x \cot x}{1 + {cosec}^2 x} dx\]
Using second fundamental theorem, evaluate the following:
`int_0^3 ("e"^x "d"x)/(1 + "e"^x)`
Evaluate the following:
`int_0^oo "e"^(-mx) x^6 "d"x`
Evaluate the following integrals as the limit of the sum:
`int_0^1 x^2 "d"x`
If `int (3"e"^x - 5"e"^-x)/(4"e"6x + 5"e"^-x)"d"x` = ax + b log |4ex + 5e –x| + C, then ______.
Verify the following:
`int (2x + 3)/(x^2 + 3x) "d"x = log|x^2 + 3x| + "C"`
Given `int "e"^"x" (("x" - 1)/("x"^2)) "dx" = "e"^"x" "f"("x") + "c"`. Then f(x) satisfying the equation is:
What is the result of a definite integral?
