Advertisements
Advertisements
प्रश्न
Evaluate the following.
`int 1/("a"^2 - "b"^2 "x"^2)` dx
Advertisements
उत्तर
Let I = `int 1/("a"^2 - "b"^2 "x"^2)` dx
`= 1/"b"^2 int 1/("a"^2/"b"^2 - "x"^2)`dx
`= 1/"b"^2 int 1/(("a"/"b")^2 - "x"^2)` dx
`= 1/"b"^2 xx 1/(2("a"/"b")) log |("a"/"b" + "x")/("a"/"b" - "x")|` + c
∴ I = `1/"2ab" log |("a" + "bx")/("a" - "bx")|` + c
Alternate Method:
Let I = `int "dx"/("a"^2 - "b"^2"x"^2) = int"dx"/("a"^2 - ("bx")^2)`
`= 1/(2 xx "a") xx 1/"b" log |("a" + "bx")/("a" - "bx")|` + c
∴ I = `1/"2ab" log |("a" + "bx")/("a" - "bx")|` + c
Notes
The answer in the textbook is incorrect.
APPEARS IN
संबंधित प्रश्न
Show that: `int1/(x^2sqrt(a^2+x^2))dx=-1/a^2(sqrt(a^2+x^2)/x)+c`
Find the particular solution of the differential equation x2dy = (2xy + y2) dx, given that y = 1 when x = 1.
Integrate the functions:
`e^(2x+3)`
Integrate the functions:
`cos x /(sqrt(1+sinx))`
Write a value of
Write a value of\[\int\frac{1}{1 + 2 e^x} \text{ dx }\].
Write a value of\[\int\frac{\sin x + \cos x}{\sqrt{1 + \sin 2x}} dx\]
Write a value of \[\int\frac{1 - \sin x}{\cos^2 x} \text{ dx }\]
The value of \[\int\frac{1}{x + x \log x} dx\] is
Integrate the following w.r.t. x : `int x^2(1 - 2/x)^2 dx`
Evaluate the following integrals:
`int x/(x + 2).dx`
Integrate the following functions w.r.t. x : `(7 + 4 + 5x^2)/(2x + 3)^(3/2)`
Integrate the following functions w.r.t. x : `(1)/(2 + 3tanx)`
Integrate the following functions w.r.t. x : `(sin6x)/(sin 10x sin 4x)`
Evaluate the following : `int (1)/(7 + 2x^2).dx`
Evaluate `int (3"x"^2 - 5)^2` dx
Evaluate the following.
`int 1/(x(x^6 + 1))` dx
If f '(x) = `1/"x" + "x"` and f(1) = `5/2`, then f(x) = log x + `"x"^2/2` + ______
State whether the following statement is True or False.
The proper substitution for `int x(x^x)^x (2log x + 1) "d"x` is `(x^x)^x` = t
State whether the following statement is True or False.
If `int x "e"^(2x)` dx is equal to `"e"^(2x)` f(x) + c, where c is constant of integration, then f(x) is `(2x - 1)/2`.
Evaluate: `int sqrt(x^2 - 8x + 7)` dx
If `int 1/(x + x^5)` dx = f(x) + c, then `int x^4/(x + x^5)`dx = ______
`int (cos2x)/(sin^2x) "d"x`
`int (x + sinx)/(1 + cosx)dx` is equal to ______.
`int sqrt(x^2 - a^2)/x dx` = ______.
The value of `sqrt(2) int (sinx dx)/(sin(x - π/4))` is ______.
Evaluate.
`int(5"x"^2 - 6"x" + 3)/(2"x" - 3) "dx"`
`int dx/((x+2)(x^2 + 1))` ...(given)
`1/(x^2 +1) dx = tan ^-1 + c`
Evaluate:
`int(sqrt(tanx) + sqrt(cotx))dx`
`int (cos4x)/(sin2x + cos2x)dx` = ______.
