# What is the indefinite integral of y=(1/x)*(7+lnx)^2 ?

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### 3 Answers

We need to find the definite integral of y = (1/x)*(7+ln x)^2.

Let u = 7 + ln x

du/dx = 1/x

=> dx / x = du

Int [ y dx] = Int [ (1/x)*(7+ln x)^2 dx ]

=> Int [ u^2 du]

=> u^3 / 3

replace u with 7 + ln x

=> (7 + ln x)^3 / 3

Therefore the integral of y = (1/x)*(7 + ln x)^2 is

**(7 + ln x)^3 / 3 + C**

To find the indefinite integral of y=(1/x)*(7+lnx)^2

We put (7+ln x) = t, then by differentiating (y+lnx) we get: (1/x) dx = dt

So thre goven integral is rewritten as:

Int y dx = Int (1/x)*(1+lnx)^2 dx = Int (7+lnx)*(1/x) dx = Int t^2 dt = (t^(2+1))/(2+1) + constant.

=> Int (1/x)*(1+lnx)^2 dx = (t^3)/3 +C

Now we replace t = (7+lnx).

**So Integral (1/x)*(1+lnx)^2 dx = (1/3)(7+lnx)^3 + C.**

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

Int f(x)dx = Int (7+ln x)^2dx/x

We'll substitute 7 + ln x = t.

We'll differentiate both sides:

dx/x = dt

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

Int t^2 dt = t^3/3 + C

But t = 7 + ln x

**The answer is: Int (7+ln x)^2dx/x = (7 + ln x)^3/3 + C**