Advertisements
Advertisements
प्रश्न
If `int (3"e"^x - 5"e"^-x)/(4"e"6x + 5"e"^-x)"d"x` = ax + b log |4ex + 5e –x| + C, then ______.
विकल्प
a = `(-1)/8`, b = `7/8`
a = `1/8`, b = `7/8`
a = `(-1)/8`, b = `(-7)/8`
a = `1/8`, b = `(-7)/8`
उत्तर
If `int (3"e"^x - 5"e"^-x)/(4"e"6x + 5"e"^-x)"d"x` = ax + b log |4ex + 5e –x| + C, then a = `(-1)/8`, b = `(-7)/8`.
Explanation:
`(3"e"^x - 5"e"^-x)/(4"e"^x + 5"e"^-x) = "a" + "b" ((4"e"^x - 5"e"^-x))/(4"e"^x + 5"e"^-x)`,
Giving 3ex – 5e –x = a(4ex + 5e–x) + b(4ex – 5e–x).
Comparing coefficients on both sides,
We get 3 = 4a + 4b and –5 = 5a – 5b.
This verifies a = `(-1)/8`, b = `7/8`.
APPEARS IN
संबंधित प्रश्न
\[\int\limits_0^{( \pi )^{2/3}} \sqrt{x} \cos^2 x^{3/2} dx\]
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
\[\int\limits_0^1 \tan^{- 1} x dx\]
\[\int\limits_0^1 \frac{1 - x}{1 + x} dx\]
\[\int\limits_0^{\pi/2} \frac{\sin x}{\sqrt{1 + \cos x}} dx\]
\[\int\limits_0^\pi \frac{x}{a^2 \cos^2 x + b^2 \sin^2 x} dx\]
\[\int\limits_1^3 \left( 2 x^2 + 5x \right) dx\]
Evaluate the following using properties of definite integral:
`int_0^(i/2) (sin^7x)/(sin^7x + cos^7x) "d"x`
Choose the correct alternative:
Γ(n) is
Evaluate `int "dx"/sqrt((x - alpha)(beta - x)), beta > alpha`