Advertisements
Advertisements
प्रश्न
Find `int x^2/(x^4 + 3x^2 + 2) "d"x`
Advertisements
उत्तर
Put x2 = t
Then 2x dx = dt.
Now I = `int (x^3"d"x)/(x^4 + 3x^2 + 2)`
= `1/2 int "tdt"/("t"^2 + 3"t" + 2)`
Consider `"t"/("t"^2 + 3"t" + 2) = "A"/("t" + 1) + "B"/("t" + 2)`
Comparing coefficient, we get A = –1, B = 2.
Then I = `1/2[2 int "dt"/("t" + 2) - int "dt"/("t" + 1)]`
= `1/2 [2log|"t" + 2| - log|"t" + 1|]`
= `log|(x^2 + 2)/sqrt(x^2 + 1)| + "C"`
APPEARS IN
संबंधित प्रश्न
If f(x) is a continuous function defined on [−a, a], then prove that
The value of \[\int\limits_0^1 \tan^{- 1} \left( \frac{2x - 1}{1 + x - x^2} \right) dx,\] is
\[\int\limits_0^{\pi/2} \frac{\sin^2 x}{\left( 1 + \cos x \right)^2} dx\]
\[\int\limits_0^{\pi/2} \frac{\cos x}{1 + \sin^2 x} dx\]
Evaluate the following integrals :-
\[\int_2^4 \frac{x^2 + x}{\sqrt{2x + 1}}dx\]
\[\int\limits_0^{\pi/2} \frac{1}{1 + \tan^3 x} dx\]
\[\int\limits_0^{\pi/2} \frac{x \sin x \cos x}{\sin^4 x + \cos^4 x} dx\]
\[\int\limits_1^4 \left( x^2 + x \right) dx\]
\[\int\limits_0^2 \left( x^2 + 2 \right) dx\]
Using second fundamental theorem, evaluate the following:
`int_1^2 (x - 1)/x^2 "d"x`
Evaluate the following integrals as the limit of the sum:
`int_1^3 (2x + 3) "d"x`
Evaluate: `int_(-1)^2 |x^3 - 3x^2 + 2x|dx`
What is the meaning of a definite integral?
Evaluate \[\int_{0}^{1}x\,dx\].
