Advertisements
Advertisements
Question
Find `int x^2/(x^4 + 3x^2 + 2) "d"x`
Advertisements
Solution
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
RELATED QUESTIONS
If \[f\left( a + b - x \right) = f\left( x \right)\] , then prove that \[\int_a^b xf\left( x \right)dx = \frac{a + b}{2} \int_a^b f\left( x \right)dx\]
Solve each of the following integral:
If \[\int\limits_0^a 3 x^2 dx = 8,\] write the value of a.
The value of \[\int\limits_0^1 \tan^{- 1} \left( \frac{2x - 1}{1 + x - x^2} \right) dx,\] is
Evaluate: \[\int\limits_{- \pi/2}^{\pi/2} \frac{\cos x}{1 + e^x}dx\] .
\[\int\limits_1^5 \frac{x}{\sqrt{2x - 1}} dx\]
\[\int\limits_0^1 \cos^{- 1} \left( \frac{1 - x^2}{1 + x^2} \right) dx\]
\[\int\limits_0^1 \sqrt{\frac{1 - x}{1 + x}} dx\]
\[\int\limits_1^2 \frac{1}{x^2} e^{- 1/x} dx\]
\[\int\limits_0^{\pi/2} \frac{x \sin x \cos x}{\sin^4 x + \cos^4 x} dx\]
\[\int\limits_0^\pi \frac{x \tan x}{\sec x + \tan x} dx\]
\[\int\limits_0^1 \cot^{- 1} \left( 1 - x + x^2 \right) dx\]
If f(x) = `{{:(x^2"e"^(-2x)",", x ≥ 0),(0",", "otherwise"):}`, then evaluate `int_0^oo "f"(x) "d"x`
Evaluate `int "dx"/sqrt((x - alpha)(beta - x)), beta > alpha`
If x = `int_0^y "dt"/sqrt(1 + 9"t"^2)` and `("d"^2y)/("d"x^2)` = ay, then a equal to ______.
The value of `int_2^3 x/(x^2 + 1)`dx is ______.
If velocity changes with time, what does a definite integral give over a time interval?
