Advertisements
Advertisements
प्रश्न
\[\int\limits_1^2 \frac{x + 3}{x\left( x + 2 \right)} dx\]
Advertisements
उत्तर
\[\int_1^2 \frac{x + 3}{x\left( x + 2 \right)} d x\]
\[ = \int_1^2 \frac{x + 2 + 1}{x\left( x + 2 \right)} d x\]
\[ = \int_1^2 \frac{1}{x}dx + \int_1^2 \frac{1}{x\left( x + 2 \right)}dx\]
\[ = \int_1^2 \frac{1}{x}dx + \frac{1}{2} \int_1^2 \frac{\left( x + 2 \right) - x}{x\left( x + 2 \right)}dx\]
\[ = \int_1^2 \frac{1}{x}dx + \frac{1}{2} \int_1^2 \frac{1}{x}dx - \frac{1}{2} \int_1^2 \frac{1}{x + 2}dx\]
\[ = \frac{3}{2} \int_1^2 \frac{1}{x}dx - \frac{1}{2} \int_1^2 \frac{1}{x + 2}dx\]
\[ = \frac{3}{2} \left[ \log x \right]_1^2 - \frac{1}{2} \left[ \log\left( x + 2 \right) \right]_1^2 \]
\[ = \frac{3}{2}\log2 - \frac{1}{2}\log4 + \frac{1}{2}\log3\]
\[ = \frac{3}{2}\log2 - \log2 + \frac{1}{2}\log3\]
\[ = \frac{1}{2}\log2 + \frac{1}{2}\log3\]
\[ = \frac{1}{2}\log6\]
APPEARS IN
संबंधित प्रश्न
\[\int_0^\frac{\pi^2}{4} \frac{\sin\sqrt{x}}{\sqrt{x}} dx\] equals
\[\int\limits_0^1 \frac{1 - x}{1 + x} dx\]
\[\int\limits_0^{\pi/3} \frac{\cos x}{3 + 4 \sin x} dx\]
Evaluate the following integrals :-
\[\int_2^4 \frac{x^2 + x}{\sqrt{2x + 1}}dx\]
\[\int\limits_0^1 x \left( \tan^{- 1} x \right)^2 dx\]
\[\int\limits_{- 1/2}^{1/2} \cos x \log\left( \frac{1 + x}{1 - x} \right) dx\]
\[\int\limits_0^{15} \left[ x^2 \right] dx\]
\[\int\limits_0^2 \left( x^2 + 2 \right) dx\]
Evaluate the following:
f(x) = `{{:("c"x",", 0 < x < 1),(0",", "otherwise"):}` Find 'c" if `int_0^1 "f"(x) "d"x` = 2
Evaluate the following using properties of definite integral:
`int_(-1)^1 log ((2 - x)/(2 + x)) "d"x`
Evaluate the following:
`int_0^oo "e"^(-4x) x^4 "d"x`
Choose the correct alternative:
The value of `int_(- pi/2)^(pi/2) cos x "d"x` is
