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
\[\int\limits_0^{( \pi )^{2/3}} \sqrt{x} \cos^2 x^{3/2} dx\]
Evaluate each of the following integral:
Evaluate the following integral:
Evaluate :
The value of \[\int\limits_0^{\pi/2} \cos x\ e^{\sin x}\ dx\] is
If \[\int\limits_0^1 f\left( x \right) dx = 1, \int\limits_0^1 xf\left( x \right) dx = a, \int\limits_0^1 x^2 f\left( x \right) dx = a^2 , then \int\limits_0^1 \left( a - x \right)^2 f\left( x \right) dx\] equals
The value of \[\int\limits_0^1 \tan^{- 1} \left( \frac{2x - 1}{1 + x - x^2} \right) dx,\] is
The value of \[\int\limits_0^{\pi/2} \log\left( \frac{4 + 3 \sin x}{4 + 3 \cos x} \right) dx\] is
Evaluate : \[\int\limits_0^\pi \frac{x}{1 + \sin \alpha \sin x}dx\] .
\[\int\limits_0^4 x\sqrt{4 - x} dx\]
\[\int\limits_0^{\pi/2} \frac{\sin^2 x}{\left( 1 + \cos x \right)^2} dx\]
\[\int\limits_1^2 \frac{1}{x^2} e^{- 1/x} dx\]
\[\int\limits_0^{\pi/4} \tan^4 x dx\]
\[\int\limits_0^1 \left| \sin 2\pi x \right| dx\]
\[\int\limits_0^\pi x \sin x \cos^4 x dx\]
Using second fundamental theorem, evaluate the following:
`int_1^2 (x - 1)/x^2 "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 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 x^3/(x + 1)` is equal to ______.
Evaluate: `int_(-1)^2 |x^3 - 3x^2 + 2x|dx`
