Advertisements
Advertisements
Question
Evaluate the following:
`int (x^2 "d"x)/((x^2 + "a"^2)(x^2 + "b"^2))`
Advertisements
Solution
Let I = `int (x^2 "d"x)/((x^2 + "a"^2)(x^2 + "b"^2))`
Put x2 = t for the purpose of partial fraction.
We get `"t"/(("t" + "a"^2)("t" + "b"^2))`
Put `"t"/(("t" + "a"^2)("t" + "b"^2)) = "A"/("T" + "a"^2) + "B"/("t" + "b"^2)`
⇒ `"t"/(("t" + "a"^2)("t" + "b"^2)) = ("A"("t" + "b"^2) + "B"("t" + "a"^2))/(("t" + "a"^2)("t" + "b"^2))`
⇒ t = At + Ab2 + Bt + Ba2
Comparing the like terms, we get
A + B = 1 and Ab2 + Ba2 = 0
A = `(-"a"^2)/"b"^2 "B"`
∴ `(-"a"^2)/"b"^2 "B" + "B"` = 1
`"B"((-"a"^2)/"b"^2 + 1)` = 1
⇒ `"B"((-"a"^2 + "b"^2)/"b"^2)` = 1
⇒ B = `"b"^2/("b"^2 - "a"^2)` and A = `(-"a"^2)/"b"^2 xx "b"^2/("b"^2 - "a"^2) = "a"^2/("a"^2 - "b"^2)`
So A = `"a"^2/("a"^2 - "b"^2)` and B = `(-"b"^2)/("a"^2 - "b"^2)`
∴ `int x^2/((x^2 + "a"^2)(x^2 + "b"^2)) "d"x = "a"^2/("a"^2 - "b"^2) int 1/(x^2 + "a"^2) "d"x - "b"^2/("a"^2 - "b"^2) int 1/(x^2 + "b"^2) "d"x`
= `"a"^2/("a"^2 - "b"^2) xx 1/"a" tan^-1 x/"a" - "b"^2/("a"^2 - "b"^2) * 1/"b" tan^-1 x/"b"`
= `"a"/("a"^2 - "b"^2) tan^-1 x/"a" - "b"/("a"^2 - "b"^2) tan^-1 x-"b" + "C"`
Hence, I = `1/("a"^2 - "b"^2) ["a" tan^-1 x/"a" - "b" tan^-1 x/"b"] + "C"`.
APPEARS IN
RELATED QUESTIONS
Integrate the rational function:
`x/((x -1)^2 (x+ 2))`
Integrate the rational function:
`2/((1-x)(1+x^2))`
Integrate the rational function:
`1/(x(x^n + 1))` [Hint: multiply numerator and denominator by xn − 1 and put xn = t]
Integrate the rational function:
`(2x)/((x^2 + 1)(x^2 + 3))`
Integrate the following w.r.t. x : `x^2/((x^2 + 1)(x^2 - 2)(x^2 + 3))`
Integrate the following w.r.t. x : `(12x^2 - 2x - 9)/((4x^2 - 1)(x + 3)`
Integrate the following w.r.t. x : `(3x - 2)/((x + 1)^2(x + 3)`
Integrate the following w.r.t. x: `(1)/(sinx + sin2x)`
Integrate the following w.r.t. x : `(1)/(sin2x + cosx)`
Integrate the following w.r.t. x : `(5*e^x)/((e^x + 1)(e^(2x) + 9)`
Choose the correct options from the given alternatives :
If `int tan^3x*sec^3x*dx = (1/m)sec^mx - (1/n)sec^n x + c, "then" (m, n)` =
Integrate the following w.r.t. x: `(x^2 + 3)/((x^2 - 1)(x^2 - 2)`
Evaluate: `int (2"x"^3 - 3"x"^2 - 9"x" + 1)/("2x"^2 - "x" - 10)` dx
`int x^7/(1 + x^4)^2 "d"x`
`int sqrt((9 + x)/(9 - x)) "d"x`
`int 1/(4x^2 - 20x + 17) "d"x`
`int "e"^x ((1 + x^2))/(1 + x)^2 "d"x`
`int (x^2 + x -1)/(x^2 + x - 6) "d"x`
`int (6x^3 + 5x^2 - 7)/(3x^2 - 2x - 1) "d"x`
`int (x + sinx)/(1 - cosx) "d"x`
`int ("d"x)/(x^3 - 1)`
Evaluate:
`int (5e^x)/((e^x + 1)(e^(2x) + 9)) dx`
`int 1/(sinx(3 + 2cosx)) "d"x`
`int xcos^3x "d"x`
Choose the correct alternative:
`int (x + 2)/(2x^2 + 6x + 5) "d"x = "p"int (4x + 6)/(2x^2 + 6x + 5) "d"x + 1/2 int 1/(2x^2 + 6x + 5)"d"x`, then p = ?
`int (5(x^6 + 1))/(x^2 + 1) "d"x` = x5 – ______ x3 + 5x + c
Evaluate `int x log x "d"x`
Evaluate `int x^2"e"^(4x) "d"x`
If `intsqrt((x - 7)/(x - 9)) dx = Asqrt(x^2 - 16x + 63) + log|x - 8 + sqrt(x^2 - 16x + 63)| + c`, then A = ______
Evaluate the following:
`int x^2/(1 - x^4) "d"x` put x2 = t
Evaluate the following:
`int (x^2"d"x)/(x^4 - x^2 - 12)`
Evaluate the following:
`int_"0"^pi (x"d"x)/(1 + sin x)`
Evaluate: `int (dx)/(2 + cos x - sin x)`
Evaluate: `int_-2^1 sqrt(5 - 4x - x^2)dx`
`int 1/(x^2 + 1)^2 dx` = ______.
Find: `int x^4/((x - 1)(x^2 + 1))dx`.
Evaluate.
`int (5x^2 - 6x + 3) / (2x -3) dx`
Evaluate:
`int(2x^3 - 1)/(x^4 + x)dx`
