How do i solve the integral Ln(x^4)/x dx?

Integrate ln(x^4)/x with respect to x.

∫{ln(x^4)/x}dx

Let

u = ln(x^4)

du = (4x³/x^4)dx = (4/x)dx

du/4 = dx/x

= ∫{ln(x^4)/x}dx = (1/4)∫u du = u²/8 + C = [ln(x^4)]²/8 + C

= [4ln(x)]²/8 + C = 2[ln(x)]² + C

integral ln (x^4) / x dx = integral 4*ln x / x

Since integral ln x /x = (ln^2 x)/2 our answer is 2 (ln^2 x)

ln(x^4) = 4 ln x.

Set u = ln x, du = 1/x dx. Then

∫ ln(x^4)/x dx

= ∫ 4u du

= 2u^2 + c

= 2 (ln x)^2 + c.

ln (x^4) / x dx = x/x^4 + c

= x / (x^2)^2

ln(x^4)/x = 4 ln x / x

let u = ln x, du = 1/x dx.

let v = x , dv = 1

∫ ln(x^4)/x dx

= 4 ∫ u/v dx

=4 [ (v ∫ u - u ∫ v) / v^2 ]

= 4[ (v x u^2/2(x)) - (u x v^2/2(1)) / v^2] + c

= 2(x)(ln x)(x) - 2(ln x)(1) + c

= 2x^2ln x - 2 ln x + c.........ans

.........gut luck......