Advertisements
Advertisements
प्रश्न
उत्तर
\[Let\ I = \int_{- 1}^1 \frac{1}{x^2 + 2x + 5} d x . Then, \]
\[ I = \int_{- 1}^1 \frac{1}{\left( x^2 + 2x + 1 \right) + 4} d x\]
\[ \Rightarrow I = \int_{- 1}^1 \frac{1}{\left( x + 1 \right)^2 + 2^2} d x\]
\[ \Rightarrow I = \frac{1}{2} \left[ \tan^{- 1} \frac{\left( x + 1 \right)}{2} \right]_{- 1}^1 \]
\[ \Rightarrow I = \frac{1}{2}\left( \tan^{- 1} 1 - \tan^{- 1} 0 \right)\]
\[ \Rightarrow I = \frac{1}{2}\left( \frac{\pi}{4} \right)\]
\[ \Rightarrow I = \frac{\pi}{8}\]
APPEARS IN
संबंधित प्रश्न
Evaluate the following definite integrals:
If \[f\left( a + b - x \right) = f\left( x \right)\] , then prove that \[\int_a^b xf\left( x \right)dx = \frac{a + b}{2} \int_a^b f\left( x \right)dx\]
Evaluate each of the following integral:
\[\int\limits_0^{\pi/2} \frac{1}{2 + \cos x} dx\] equals
Given that \[\int\limits_0^\infty \frac{x^2}{\left( x^2 + a^2 \right)\left( x^2 + b^2 \right)\left( x^2 + c^2 \right)} dx = \frac{\pi}{2\left( a + b \right)\left( b + c \right)\left( c + a \right)},\] the value of \[\int\limits_0^\infty \frac{dx}{\left( x^2 + 4 \right)\left( x^2 + 9 \right)},\]
\[\int\limits_0^1 \cos^{- 1} \left( \frac{1 - x^2}{1 + x^2} \right) dx\]
\[\int\limits_0^{\pi/3} \frac{\cos x}{3 + 4 \sin x} dx\]
\[\int\limits_0^{\pi/4} \tan^4 x dx\]
\[\int\limits_{- 1/2}^{1/2} \cos x \log\left( \frac{1 + x}{1 - x} \right) dx\]
\[\int\limits_1^3 \left( 2 x^2 + 5x \right) dx\]
Using second fundamental theorem, evaluate the following:
`int_1^2 (x "d"x)/(x^2 + 1)`
Evaluate the following using properties of definite integral:
`int_0^1 log (1/x - 1) "d"x`
Choose the correct alternative:
Γ(n) is
Integrate `((2"a")/sqrt(x) - "b"/x^2 + 3"c"root(3)(x^2))` w.r.t. x
Evaluate `int (x^2 + x)/(x^4 - 9) "d"x`
If `int (3"e"^x - 5"e"^-x)/(4"e"6x + 5"e"^-x)"d"x` = ax + b log |4ex + 5e –x| + C, then ______.
Given `int "e"^"x" (("x" - 1)/("x"^2)) "dx" = "e"^"x" "f"("x") + "c"`. Then f(x) satisfying the equation is:
Find: `int logx/(1 + log x)^2 dx`