Advertisements
Advertisements
प्रश्न
Advertisements
उत्तर
\[Let\ I = \int_1^2 \frac{x}{\left( x + 1 \right)\left( x + 2 \right)} d\ x\ . Then, \]
\[I = \int_1^2 \left( \frac{- 1}{\left( x + 1 \right)} + \frac{2}{\left( x + 2 \right)} \right) d x\]
\[ \Rightarrow I = - \int_1^2 \frac{1}{\left( x + 1 \right)} dx + 2 \int_1^2 \frac{1}{\left( x + 2 \right)} dx\]
\[ \Rightarrow I = \left[ - \log \left( x + 1 \right) + 2 \log \left( x + 2 \right) \right]_1^2 \]
\[ \Rightarrow I = - \log 3 + 2 \log 4 + \log 2 - 2 \log 3\]
\[ \Rightarrow I = 5 \log 2 - 3 \log 3\]
\[ \Rightarrow I = \log 2^5 - \log 3^3 \]
\[ \Rightarrow I = \log \frac{32}{27}\]
APPEARS IN
संबंधित प्रश्न
If f (x) is a continuous function defined on [0, 2a]. Then, prove that
\[\int_0^\frac{\pi^2}{4} \frac{\sin\sqrt{x}}{\sqrt{x}} dx\] equals
\[\int\limits_0^{2a} f\left( x \right) dx\] is equal to
Evaluate : \[\int\limits_0^\pi \frac{x}{1 + \sin \alpha \sin x}dx\] .
Evaluate: \[\int\limits_{- \pi/2}^{\pi/2} \frac{\cos x}{1 + e^x}dx\] .
\[\int\limits_0^{1/\sqrt{3}} \tan^{- 1} \left( \frac{3x - x^3}{1 - 3 x^2} \right) dx\]
\[\int\limits_0^{\pi/4} \tan^4 x dx\]
\[\int\limits_2^3 e^{- x} dx\]
Prove that `int_a^b ƒ ("x") d"x" = int_a^bƒ(a + b - "x") d"x" and "hence evaluate" int_(π/6)^(π/3) (d"x")/(1+sqrt(tan "x")`
Evaluate the following integrals as the limit of the sum:
`int_1^3 (2x + 3) "d"x`
Choose the correct alternative:
`int_(-1)^1 x^3 "e"^(x^4) "d"x` is
Choose the correct alternative:
If n > 0, then Γ(n) is
Given `int "e"^"x" (("x" - 1)/("x"^2)) "dx" = "e"^"x" "f"("x") + "c"`. Then f(x) satisfying the equation is:
