Advertisements
Advertisements
Question
Advertisements
Solution
\[Let\ I = \int_0^1 \frac{1 - x^2}{\left( 1 + x^2 \right)^2} d\ x . Then, \]
\[I = \int_0^1 \frac{\left( \frac{1}{x^2} - 1 \right)}{\left( x + \frac{1}{x} \right)^2} dx\]
\[Let\ x + \frac{1}{x} = t . Then, 1 - \frac{1}{x^2} dx = dt\]
\[When\ x = 0, t\ = \infty\ and\ x\ = 1, t = 2\]
\[ \therefore I = \int_\infty^2 \frac{- dt}{t^2}\]
\[ \Rightarrow I = \left[ \frac{1}{t} \right]_\infty^2 \]
\[ \Rightarrow I = \frac{1}{2} - 0\]
\[ \Rightarrow I = \frac{1}{2}\]
APPEARS IN
RELATED QUESTIONS
Evaluate each of the following integral:
\[\int_a^b \frac{x^\frac{1}{n}}{x^\frac{1}{n} + \left( a + b - x \right)^\frac{1}{n}}dx, n \in N, n \geq 2\]
\[\int\limits_0^\pi \frac{1}{1 + \sin x} dx\] equals
The value of \[\int\limits_{- \pi/2}^{\pi/2} \left( x^3 + x \cos x + \tan^5 x + 1 \right) dx, \] is
Evaluate : \[\int\limits_0^\pi/4 \frac{\sin x + \cos x}{16 + 9 \sin 2x}dx\] .
\[\int\limits_0^{\pi/2} \frac{\sin^2 x}{\left( 1 + \cos x \right)^2} dx\]
\[\int\limits_0^1 \left( \cos^{- 1} x \right)^2 dx\]
\[\int\limits_0^\pi \frac{x \sin x}{1 + \cos^2 x} dx\]
\[\int\limits_0^{\pi/2} \frac{x \sin x \cos x}{\sin^4 x + \cos^4 x} dx\]
\[\int\limits_{- \pi}^\pi x^{10} \sin^7 x dx\]
\[\int\limits_0^{\pi/2} \frac{1}{2 \cos x + 4 \sin x} dx\]
\[\int\limits_1^3 \left( 2 x^2 + 5x \right) dx\]
Using second fundamental theorem, evaluate the following:
`int_0^(1/4) sqrt(1 - 4) "d"x`
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_(- pi/4)^(pi/4) x^3 cos^3 x "d"x`
If f(x) = `{{:(x^2"e"^(-2x)",", x ≥ 0),(0",", "otherwise"):}`, then evaluate `int_0^oo "f"(x) "d"x`
Choose the correct alternative:
If n > 0, then Γ(n) is
Choose the correct alternative:
`int_0^oo x^4"e"^-x "d"x` is
Integrate `((2"a")/sqrt(x) - "b"/x^2 + 3"c"root(3)(x^2))` w.r.t. x
Find: `int logx/(1 + log x)^2 dx`
