Advertisements
Advertisements
प्रश्न
Evaluate `int (x^2"d"x)/(x^4 + x^2 - 2)`
Advertisements
उत्तर
Let x2 = t.
Then `x^2/(x^4 + x^2 - 2) = "t"/("t"^2 + "t" - 2)`
= `"t"/(("t" + 2)("t" - 1))`
= `"A"/("t" + 2) + "B"/("t" - 1)`
So t = A(t – 1) + B(t + 2)
Comparing coefficients, we get A = `2/3`, B = `1/3`.
So `x^2/(x^4 + x^2 - 2) = 2/3 1/(x^2 + 2) + 1/3 1/(x^2 - 1)`
Therefore, `int x^2/(x^4 + x^2 - 2) "d"x`
= `2/3 int 1/(x^2 + 2) "d"x + 1/3 int "dx"/(x^2 - 1)`
= `2/3 1/sqrt(2) tan^-1 x/sqrt(2) + 1/6 log |(x + 1)/(x + 1)| + "C"`
APPEARS IN
संबंधित प्रश्न
Evaluate the following integral:
If f(2a − x) = −f(x), prove that
If f (x) is a continuous function defined on [0, 2a]. Then, prove that
The value of the integral \[\int\limits_0^\infty \frac{x}{\left( 1 + x \right)\left( 1 + x^2 \right)} dx\]
If \[I_{10} = \int\limits_0^{\pi/2} x^{10} \sin x\ dx,\] then the value of I10 + 90I8 is
\[\int\limits_0^{2a} f\left( x \right) dx\] is equal to
\[\int\limits_0^4 x\sqrt{4 - x} dx\]
\[\int\limits_0^{\pi/2} \frac{\sin x}{\sqrt{1 + \cos x}} dx\]
\[\int\limits_0^{\pi/4} \sin 2x \sin 3x dx\]
\[\int\limits_0^{\pi/2} \frac{1}{1 + \tan^3 x} dx\]
\[\int\limits_0^\pi \frac{x}{a^2 \cos^2 x + b^2 \sin^2 x} dx\]
\[\int\limits_0^\pi \frac{x \tan x}{\sec x + \tan x} dx\]
\[\int\limits_0^\pi \frac{dx}{6 - \cos x}dx\]
\[\int\limits_2^3 e^{- x} dx\]
Using second fundamental theorem, evaluate the following:
`int_0^1 x"e"^(x^2) "d"x`
Choose the correct alternative:
`int_0^1 (2x + 1) "d"x` is
Choose the correct alternative:
Γ(1) is
If x = `int_0^y "dt"/sqrt(1 + 9"t"^2)` and `("d"^2y)/("d"x^2)` = ay, then a equal to ______.
Verify the following:
`int (x - 1)/(2x + 3) "d"x = x - log |(2x + 3)^2| + "C"`
