Advertisements
Advertisements
प्रश्न
Evaluate the following.
`int "x"^3/(16"x"^8 - 25)` dx
Advertisements
उत्तर
Let I = `int "x"^3/(16"x"^8 - 25)` dx
Put x4 = t
∴ 4x3 dx = dt
∴ x3 dx = `1/4` dt
∴ I = `1/4 int "dt"/(16"t"^2 - 25)`
`= 1/(4 xx 16) int "dt"/("t"^2 - 25/16)`
`= 1/64 int "dt"/("t"^2 - (5/4)^2)`
`= 1/64 xx 1/(2 xx 5/4) log |("t" - 5/4)/("t" + 5/4)|` + c
`= 1/160 log |("4t" - 5)/("4t" + 5)|` + c
∴ I = = `1/160 log |(4"x"^4 - 5)/(4"x"^4 + 5)|` + c
Notes
The answer in the textbook is incorrect.
APPEARS IN
संबंधित प्रश्न
Integrate the functions:
`(sin^(-1) x)/(sqrt(1-x^2))`
Integrate the functions:
`cos x /(sqrt(1+sinx))`
Evaluate : `∫1/(3+2sinx+cosx)dx`
Write a value of
Write a value of\[\int\left( e^{x \log_e \text{ a}} + e^{a \log_e x} \right) dx\] .
Evaluate the following integral:
`int(4x + 3)/(2x + 1).dx`
Integrate the following functions w.r.t. x : sin4x.cos3x
Integrate the following functions w.r.t. x : `(7 + 4 + 5x^2)/(2x + 3)^(3/2)`
Evaluate the following : `int (1)/(x^2 + 8x + 12).dx`
Choose the correct options from the given alternatives :
`int f x^x (1 + log x)*dx`
Integrate the following w.r.t.x: `(3x + 1)/sqrt(-2x^2 + x + 3)`
Evaluate `int (3"x"^3 - 2sqrt"x")/"x"` dx
Evaluate: `int "e"^sqrt"x"` dx
`int cos sqrtx` dx = _____________
`int "e"^x[((x + 3))/((x + 4)^2)] "d"x`
`int x^x (1 + logx) "d"x`
`int (cos2x)/(sin^2x) "d"x`
`int(1 - x)^(-2) dx` = ______.
`int "e"^(sin^-1 x) ((x + sqrt(1 - x^2))/(sqrt1 - x^2)) "dx" = ?`
`int_1^3 ("d"x)/(x(1 + logx)^2)` = ______.
`int sqrt(x^2 - a^2)/x dx` = ______.
Evaluate the following.
`int x^3/(sqrt(1+x^4))dx`
Evaluate the following.
`int(20 - 12"e"^"x")/(3"e"^"x" - 4) "dx"`
Evaluate the following.
`int 1/(x^2 + 4x - 5)dx`
Evaluate the following.
`int x sqrt(1 + x^2) dx`
`int 1/(sin^2x cos^2x)dx` = ______.
Evaluate `int 1/(x(x-1))dx`
Evaluate the following.
`intx^3/sqrt(1 + x^4)dx`
What is integration by substitution?
