i still need help on this...integrate (-2)/(xln^4(x) dx...should i be using substitution ...if yes then wut should i substitute? plz help me..thanks in advanced!
at is wrong with u= 1/x and dv= dx/ln^4(x) ?
remember d/dx (lnx)^-n= (-n/x)*lnx)^(n-1)
okey thank you...i got it now
To integrate the function (-2)/(xln^4(x))dx, you can use integration by parts. The standard formula for integration by parts is:
∫u dv = uv - ∫v du
In this case, let's choose:
u = -2
dv = (1/(xln^4(x)))dx
To determine du and v, we need to differentiate u and integrate dv.
Differentiating u gives:
du = 0 dx = 0
Integrating dv can be a bit tricky, but we can simplify the integrand first using a substitution.
Let's make the substitution u = ln(x). Then, we have:
du = (1/x)dx
The integral then becomes:
∫(xln^4(x))^(-1) (1/x)dx
Simplifying further, we get:
∫(u^4)^(−1) du = ∫u^(-4) du
Integrating ∫u^(-4) du, we apply the formula:
∫x^n dx = (x^(n+1))/(n+1) + C
Using this formula, the integral becomes:
∫u^(-4) du = (u^(-4 + 1))/(-4 + 1) + C
= (u^(-3))/(-3) + C
= -1/(3u^3) + C
= -1/(3(ln(x))^3) + C
Now we can go back to the formula for integration by parts:
∫u dv = uv - ∫v du
Substituting the values back in, we have:
∫(-2) dx/(xln^4(x)) = -2 * (-1/(3(ln(x))^3)) - ∫(-1/(3(ln(x))^3)) * 0 dx
= 2/(3(ln(x))^3) + C
Thus, the final result of the integral is:
∫(-2)/(xln^4(x))dx = 2/(3(ln(x))^3) + C