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:
If \[\left[ \cdot \right] and \left\{ \cdot \right\}\] denote respectively the greatest integer and fractional part functions respectively, evaluate the following integrals:
If \[I_{10} = \int\limits_0^{\pi/2} x^{10} \sin x\ dx,\] then the value of I10 + 90I8 is
The value of the integral \[\int\limits_{- 2}^2 \left| 1 - x^2 \right| dx\] is ________ .
Evaluate : \[\int\limits_0^\pi \frac{x}{1 + \sin \alpha \sin x}dx\] .
\[\int\limits_1^3 \left| x^2 - 2x \right| dx\]
\[\int\limits_1^3 \left| x^2 - 4 \right| dx\]
\[\int\limits_0^\pi \frac{x \sin x}{1 + \cos^2 x} dx\]
\[\int\limits_0^{\pi/2} \frac{x \sin x \cos x}{\sin^4 x + \cos^4 x} dx\]
\[\int\limits_2^3 \frac{\sqrt{x}}{\sqrt{5 - x} + \sqrt{x}} dx\]
Using second fundamental theorem, evaluate the following:
`int_0^(1/4) sqrt(1 - 4) "d"x`
Using second fundamental theorem, evaluate the following:
`int_0^1 x"e"^(x^2) "d"x`
Evaluate the following:
`int_0^oo "e"^(-4x) x^4 "d"x`
Choose the correct alternative:
`int_(-1)^1 x^3 "e"^(x^4) "d"x` is
Evaluate `int (3"a"x)/("b"^2 + "c"^2x^2) "d"x`
Find `int x^2/(x^4 + 3x^2 + 2) "d"x`
If `int (3"e"^x - 5"e"^-x)/(4"e"6x + 5"e"^-x)"d"x` = ax + b log |4ex + 5e –x| + C, then ______.
`int "e"^x ((1 - x)/(1 + x^2))^2 "d"x` is equal to ______.
