Advertisements
Advertisements
प्रश्न
Advertisements
उत्तर
\[Let\ x^2 = t . Then, 2x\ dx = dt\]
\[When\ x = 1, t = 1\ and\ x = 2, t = 4\]
\[ \therefore I = \int_1^2 \frac{3x}{9 x^2 - 1} d x\]
\[ \Rightarrow I = \frac{3}{2} \int_1^4 \frac{dt}{9t - 1}\]
\[ \Rightarrow I = \frac{3}{18} \left[ \log \left( 9t - 1 \right) \right]_1^4 \]
\[ \Rightarrow I = \frac{3}{18}\left( \log 35 - \log 8 \right)\]
\[ \Rightarrow I = \frac{\left( \log 35 - \log 8 \right)}{6}\]
APPEARS IN
संबंधित प्रश्न
Evaluate the following integral:
If f (x) is a continuous function defined on [0, 2a]. Then, prove that
Evaluate each of the following integral:
The value of the integral \[\int\limits_0^{\pi/2} \frac{\sqrt{\cos x}}{\sqrt{\cos x} + \sqrt{\sin x}} dx\] is
\[\int\limits_0^\infty \frac{1}{1 + e^x} dx\] equals
\[\int_0^\frac{\pi^2}{4} \frac{\sin\sqrt{x}}{\sqrt{x}} dx\] equals
The value of the integral \[\int\limits_0^\infty \frac{x}{\left( 1 + x \right)\left( 1 + x^2 \right)} dx\]
If \[\int\limits_0^a \frac{1}{1 + 4 x^2} dx = \frac{\pi}{8},\] then a equals
Evaluate : \[\int\limits_0^{2\pi} \cos^5 x dx\] .
\[\int\limits_0^1 \cos^{- 1} x dx\]
\[\int\limits_0^{1/\sqrt{3}} \tan^{- 1} \left( \frac{3x - x^3}{1 - 3 x^2} \right) dx\]
\[\int\limits_0^1 \frac{1 - x}{1 + x} dx\]
\[\int\limits_0^{\pi/2} \frac{\sin^2 x}{\left( 1 + \cos x \right)^2} dx\]
\[\int\limits_1^3 \left| x^2 - 2x \right| dx\]
\[\int\limits_0^\pi \frac{x}{a^2 - \cos^2 x} dx, a > 1\]
Find : `∫_a^b logx/x` dx
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`
Integrate `((2"a")/sqrt(x) - "b"/x^2 + 3"c"root(3)(x^2))` w.r.t. x
Find `int sqrt(10 - 4x + 4x^2) "d"x`
The value of `int_2^3 x/(x^2 + 1)`dx is ______.
