# Evaluate the integral of the function y=(4+ln x)^3/x?

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We have to find the integral of y=(4+ln x)^3/x.

To start, let us take t = ln x

=> dt/dx = 1/x

=> dx/x = dt

Int [y] = Int [(4+ln x)^3/x dx]

=> Int [ (dx/x) ( 4 + ln x)]

=> Int [ (4 + t)^3 dt]

=> [(4 + t)^4]/ 4 + C

replace t with ln x

=> [ 4 + ln x]^4 / 4 + C

**Therefore the integral of y=(4+ln x)^3/x is (4 + ln x)^4 / 4 + C**

we can raghit to solve thes integral:

Int f(x)=(4+lnX)^3/x (dx) = Int (4+ln x)^3dx/x

We'll substitute 4 + ln x = t.

dx/x = dt

We'll re-write the integral :

Int t^3 dt = t^4/4 + C

But t = 4 + ln x

**Int (4+ln x)^3dx/x = (4 + ln x)^4/4 + C**

We'll apply the substitution technique to solve the indefinite integral of the given function.

Int f(x)dx = Int (4+ln x)^3dx/x

We'll substitute 4 + ln x = t.

We'll differentiate both sides:

dx/x = dt

We'll re-write the integral, having t as variable:

Int t^3 dt = t^4/4 + C

But t = 4 + ln x

**Int (4+ln x)^3dx/x = (4 + ln x)^4/4 + C**