Advertisements
Advertisements
प्रश्न
Verify the following:
`int (x - 1)/(2x + 3) "d"x = x - log |(2x + 3)^2| + "C"`
Advertisements
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
Evaluate the following integral:
Evaluate each of the following integral:
\[\int\limits_0^1 \left\{ x \right\} dx,\] where {x} denotes the fractional part of x.
\[\int\limits_0^{2a} f\left( x \right) dx\] is equal to
`int_0^(2a)f(x)dx`
\[\int\limits_0^1 \cos^{- 1} \left( \frac{1 - x^2}{1 + x^2} \right) dx\]
\[\int\limits_0^1 \tan^{- 1} \left( \frac{2x}{1 - x^2} \right) dx\]
\[\int\limits_0^1 \log\left( 1 + x \right) dx\]
\[\int\limits_1^3 \left| x^2 - 4 \right| dx\]
\[\int\limits_{- a}^a \frac{x e^{x^2}}{1 + x^2} dx\]
\[\int\limits_0^1 \cot^{- 1} \left( 1 - x + x^2 \right) dx\]
\[\int\limits_{\pi/6}^{\pi/2} \frac{\ cosec x \cot x}{1 + {cosec}^2 x} dx\]
\[\int\limits_1^4 \left( x^2 + x \right) dx\]
\[\int\limits_2^3 e^{- x} dx\]
Using second fundamental theorem, evaluate the following:
`int_0^(1/4) sqrt(1 - 4) "d"x`
Evaluate the following:
`int_(-1)^1 "f"(x) "d"x` where f(x) = `{{:(x",", x ≥ 0),(-x",", x < 0):}`
Choose the correct alternative:
If n > 0, then Γ(n) is
Evaluate `int (3"a"x)/("b"^2 + "c"^2x^2) "d"x`
Given `int "e"^"x" (("x" - 1)/("x"^2)) "dx" = "e"^"x" "f"("x") + "c"`. Then f(x) satisfying the equation is:
