Advertisements
Advertisements
प्रश्न
Evaluate the following.
`int 1/(x(x^6 + 1))` dx
Advertisements
उत्तर
Let I = `int 1/(x(x^6 + 1))` dx
`= int x^5/(x^6(x^6 + 1))`dx
Put x6 = t
∴ 6x5 dx = dt
∴ `x^5 * dx = 1/6 * dt`
∴ I = `1/6 int dt/(t(t + 1))`
`= 1/6 int ((t + 1) - t)/(t(t + 1))` dt
`= 1/6 int (1/t - 1/(t + 1))` dt
= `1/6` [log | t | - log |t + 1|] + c
`= 1/6 log |t/(t + 1)|` + c
∴ I = `1/6 log |x^6/(x^6 + 1)|` + c
संबंधित प्रश्न
Integrate the functions:
`e^(tan^(-1)x)/(1+x^2)`
Solve:
dy/dx = cos(x + y)
Write a value of
Write a value of\[\int a^x e^x \text{ dx }\]
Write a value of\[\int\frac{\sin x - \cos x}{\sqrt{1 + \sin 2x}} \text{ dx}\]
Write a value of\[\int e^{ax} \sin\ bx\ dx\]
Find : ` int (sin 2x ) /((sin^2 x + 1) ( sin^2 x + 3 ) ) dx`
Integrate the following functions w.r.t. x : `(2x + 1)sqrt(x + 2)`
Evaluate the following : `int sqrt((9 + x)/(9 - x)).dx`
Integrate the following with respect to the respective variable:
`x^7/(x + 1)`
Evaluate the following.
`int 1/(sqrt"x" + "x")` dx
Evaluate the following.
`int 1/(sqrt(3"x"^2 - 5))` dx
Fill in the Blank.
`int (5("x"^6 + 1))/("x"^2 + 1)` dx = x4 + ______ x3 + 5x + c
Evaluate: `int sqrt(x^2 - 8x + 7)` dx
`int x^2/sqrt(1 - x^6)` dx = ________________
`int (2 + cot x - "cosec"^2x) "e"^x "d"x`
`int(5x + 2)/(3x - 4) dx` = ______
If I = `int (sin2x)/(3x + 4cosx)^3 "d"x`, then I is equal to ______.
`int[ tan (log x) + sec^2 (log x)] dx= ` ______
`int 1/(a^2 - x^2) dx = 1/(2a) xx` ______.
`int(log(logx) + 1/(logx)^2)dx` = ______.
`int (x + sinx)/(1 + cosx)dx` is equal to ______.
`int dx/(2 + cos x)` = ______.
(where C is a constant of integration)
Find `int (x + 2)/sqrt(x^2 - 4x - 5) dx`.
Evaluate `int(1 + x + x^2/(2!) )dx`
Evaluate:
`int(sqrt(tanx) + sqrt(cotx))dx`
Evaluate `int(5x^2-6x+3)/(2x-3)dx`
