Advertisements
Advertisements
प्रश्न
Evaluate `int (x^2 + x)/(x^4 - 9) "d"x`
Advertisements
उत्तर
We have I = `int (x^2 + x)/(x^4 - 9) "d"x`
= `int x^2/(x^4 - 9) "d"x + (x"d"x)/(x^4 - 9)`
= I1 + I2
Now I1 = int x^3/(x^4 - 9)`
Put t = x4 – 9
So that 4x3 dx = dt.
Therefore I1 = `1/4 int "dt"/"t"`
= `1/4 log|"t"| + "C"_1`
= `1/4 log|x^4 - 9| + "C"_1`
Again, I2 = `int (x"d"x)/(x^4 - 9)`
Put x2 = u
So that 2x dx = du
Then I2 = `1/2 int "du"/("u"^2 - (3)^2)`
= `1/(2 xx 6) log|("u" - 3)/("u" + 3)| + "C"_2`
= `1/12 log|(x^2 - 3)/(x^2 + 3)| + "C"_2`.
Thus I = I1 + I2
= `1/4 log|x^4 - 9| + 1/12 log|(x^2 - 3)/(x^2 + 3)| + "C"`
APPEARS IN
संबंधित प्रश्न
Evaluate each of the following integral:
The value of \[\int\limits_0^{2\pi} \sqrt{1 + \sin\frac{x}{2}}dx\] is
\[\int\limits_0^{2a} f\left( x \right) dx\] is equal to
\[\int\limits_1^2 x\sqrt{3x - 2} dx\]
\[\int\limits_0^{\pi/2} \frac{\cos x}{1 + \sin^2 x} dx\]
\[\int\limits_0^1 \left( \cos^{- 1} x \right)^2 dx\]
\[\int\limits_0^1 \left| 2x - 1 \right| dx\]
\[\int\limits_0^{\pi/2} \left| \sin x - \cos x \right| dx\]
\[\int\limits_0^{2\pi} \cos^7 x dx\]
\[\int\limits_0^\pi \frac{x}{a^2 \cos^2 x + b^2 \sin^2 x} dx\]
\[\int\limits_{- \pi/4}^{\pi/4} \left| \tan x \right| dx\]
\[\int\limits_0^{\pi/2} \frac{x \sin x \cos x}{\sin^4 x + \cos^4 x} dx\]
\[\int\limits_1^3 \left( 2 x^2 + 5x \right) dx\]
Evaluate the following:
`int_(-1)^1 "f"(x) "d"x` where f(x) = `{{:(x",", x ≥ 0),(-x",", x < 0):}`
Choose the correct alternative:
If f(x) is a continuous function and a < c < b, then `int_"a"^"c" f(x) "d"x + int_"c"^"b" f(x) "d"x` is
`int "e"^x ((1 - x)/(1 + x^2))^2 "d"x` is equal to ______.
`int x^9/(4x^2 + 1)^6 "d"x` is equal to ______.
Evaluate: `int_(-1)^2 |x^3 - 3x^2 + 2x|dx`
