Advertisements
Advertisements
Question
Advertisements
Solution
\[Let\ I = \int_1^2 \frac{x}{\left( x + 1 \right)\left( x + 2 \right)} d\ x\ . Then, \]
\[I = \int_1^2 \left( \frac{- 1}{\left( x + 1 \right)} + \frac{2}{\left( x + 2 \right)} \right) d x\]
\[ \Rightarrow I = - \int_1^2 \frac{1}{\left( x + 1 \right)} dx + 2 \int_1^2 \frac{1}{\left( x + 2 \right)} dx\]
\[ \Rightarrow I = \left[ - \log \left( x + 1 \right) + 2 \log \left( x + 2 \right) \right]_1^2 \]
\[ \Rightarrow I = - \log 3 + 2 \log 4 + \log 2 - 2 \log 3\]
\[ \Rightarrow I = 5 \log 2 - 3 \log 3\]
\[ \Rightarrow I = \log 2^5 - \log 3^3 \]
\[ \Rightarrow I = \log \frac{32}{27}\]
APPEARS IN
RELATED QUESTIONS
Evaluate the following definite integrals:
\[\int\limits_0^{( \pi )^{2/3}} \sqrt{x} \cos^2 x^{3/2} dx\]
If \[f\left( a + b - x \right) = f\left( x \right)\] , then prove that \[\int_a^b xf\left( x \right)dx = \frac{a + b}{2} \int_a^b f\left( x \right)dx\]
If f is an integrable function, show that
\[\int\limits_0^\infty \frac{1}{1 + e^x} dx\] equals
The value of \[\int\limits_0^{\pi/2} \cos x\ e^{\sin x}\ dx\] is
If \[I_{10} = \int\limits_0^{\pi/2} x^{10} \sin x\ dx,\] then the value of I10 + 90I8 is
\[\int\limits_0^4 x\sqrt{4 - x} dx\]
\[\int\limits_0^{\pi/4} \sin 2x \sin 3x dx\]
\[\int\limits_1^2 \frac{1}{x^2} e^{- 1/x} dx\]
Evaluate the following integrals :-
\[\int_2^4 \frac{x^2 + x}{\sqrt{2x + 1}}dx\]
\[\int\limits_0^1 \left| 2x - 1 \right| dx\]
\[\int\limits_0^{\pi/2} \left| \sin x - \cos x \right| dx\]
\[\int\limits_{\pi/6}^{\pi/2} \frac{\ cosec x \cot x}{1 + {cosec}^2 x} dx\]
\[\int\limits_1^4 \left( x^2 + x \right) dx\]
\[\int\limits_1^3 \left( x^2 + 3x \right) dx\]
Using second fundamental theorem, evaluate the following:
`int_1^"e" ("d"x)/(x(1 + logx)^3`
Evaluate the following:
`int_0^2 "f"(x) "d"x` where f(x) = `{{:(3 - 2x - x^2",", x ≤ 1),(x^2 + 2x - 3",", 1 < x ≤ 2):}`
Evaluate the following:
`int_(-1)^1 "f"(x) "d"x` where f(x) = `{{:(x",", x ≥ 0),(-x",", x < 0):}`
Choose the correct alternative:
If f(x) is a continuous function and a < c < b, then `int_"a"^"c" f(x) "d"x + int_"c"^"b" f(x) "d"x` is
Choose the correct alternative:
Γ(1) is
Evaluate `int "dx"/sqrt((x - alpha)(beta - x)), beta > alpha`
If `int (3"e"^x - 5"e"^-x)/(4"e"6x + 5"e"^-x)"d"x` = ax + b log |4ex + 5e –x| + C, then ______.
