Advertisements
Advertisements
प्रश्न
Evaluate the following.
`int 1/(x(x^6 + 1))` dx
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:
`(sin x)/(1+ cos x)^2`
Integrate the functions:
`1/(1 + cot x)`
Write a value of
Write a value of\[\int\frac{\sin x + \cos x}{\sqrt{1 + \sin 2x}} dx\]
Integrate the following w.r.t. x : `int x^2(1 - 2/x)^2 dx`
Integrate the following functions w.r.t. x : `(logx)^n/x`
Integrate the following functions w.r.t. x : `e^(3x)/(e^(3x) + 1)`
Integrate the following functions w.r.t.x:
`(2sinx cosx)/(3cos^2x + 4sin^2 x)`
Evaluate the following integrals:
`int (2x + 1)/(x^2 + 4x - 5).dx`
Choose the correct option from the given alternatives :
`int (1 + x + sqrt(x + x^2))/(sqrt(x) + sqrt(1 + x))*dx` =
Evaluate: `int sqrt("x"^2 + 2"x" + 5)` dx
`int (2(cos^2 x - sin^2 x))/(cos^2 x + sin^2 x)` dx = ______________
`int cos^7 x "d"x`
`int sqrt(("e"^(3x) - "e"^(2x))/("e"^x + 1)) "d"x`
`int(log(logx) + 1/(logx)^2)dx` = ______.
`int x/sqrt(1 - 2x^4) dx` = ______.
(where c is a constant of integration)
`int(1 - x)^(-2)` dx = `(1 - x)^(-1) + c`
`int secx/(secx - tanx)dx` equals ______.
if `f(x) = 4x^3 - 3x^2 + 2x +k, f (0) = - 1 and f (1) = 4, "find " f(x)`
`int dx/((x+2)(x^2 + 1))` ...(given)
`1/(x^2 +1) dx = tan ^-1 + c`
Prove that:
`int 1/sqrt(x^2 - a^2) dx = log |x + sqrt(x^2 - a^2)| + c`.
Evaluate:
`int 1/(1 + cosα . cosx)dx`
Evaluate the following.
`int(1)/(x^2 + 4x - 5)dx`
Evaluate `int (1 + x + x^2/(2!)) dx`
Evaluate the following.
`int (x^3)/(sqrt(1 + x^4)) dx`
Evaluate the following.
`int "x"^3/sqrt(1 + "x"^4)` dx
Evaluate `int 1/(x(x-1))dx`
Evaluate the following.
`int1/(x^2 + 4x - 5)dx`