Advertisements
Advertisements
प्रश्न
Verify the following:
`int (x - 1)/(2x + 3) "d"x = x - log |(2x + 3)^2| + "C"`
उत्तर
L.H.S. = `int (2x - 1)/(2x + 3) "d"x`
⇒ `int (1 - 4/(2x + 3)) "d"x` .....[Dividing the numerator by the denominator]
⇒ `int 1 * "d"x - 4 int 1/(2x + 3) "d"x`
⇒ `int 1 * "d"x - 4/2 int 1/(x + 3/2) "d"x`
⇒ `int 1 * "d"x - 2 int 1/(x + 3/2) "d"x`
⇒ `x - 2 log |x + 3/2| + "C"`
⇒ `x - 2 log |(2x + 3)/2| + "C"`
⇒ `x - log|((2x + 3)/2)^2| + "C"` ....[∵ n log m = log mn]
⇒ `x - log |(2x + 3)^2| - log 2^2 + "C"`
⇒ `x - log |(2x + 3)^2| + "C"_1`
⇒ R.H.S. ......[Where C1 = C – log 22]
L.H.S. = R.H.S.
Hence proved.
APPEARS IN
संबंधित प्रश्न
\[\int\limits_0^\infty \frac{1}{1 + e^x} dx\] equals
\[\int\limits_0^{\pi/4} \sin 2x \sin 3x dx\]
\[\int\limits_0^1 \sqrt{\frac{1 - x}{1 + x}} dx\]
\[\int\limits_1^2 \frac{x + 3}{x\left( x + 2 \right)} dx\]
\[\int\limits_0^a \frac{\sqrt{x}}{\sqrt{x} + \sqrt{a - x}} dx\]
\[\int\limits_0^{\pi/2} \frac{1}{1 + \tan^3 x} dx\]
Using second fundamental theorem, evaluate the following:
`int_0^1 x"e"^(x^2) "d"x`
Evaluate the following:
`int_0^oo "e"^(-mx) x^6 "d"x`
Choose the correct alternative:
The value of `int_(- pi/2)^(pi/2) cos x "d"x` is
Choose the correct alternative:
Using the factorial representation of the gamma function, which of the following is the solution for the gamma function Γ(n) when n = 8 is
