This may look complicated at first but it can be broken down into a number of simple steps...

For the left side of the equation, the d/dx tells you we are differentiating and since (x³ + 3x²)ln(x) sees two functions multiplied together, we therefore use the product rule.

The product rule looks like this:

u v' + u' v

where in this case u = (x³ + 3x²) and v = ln(x)

u' is the differential of u. Differentiating here simply involves multiplying the coefficient (number in front of the x) by the power and then subtracting one from the power so:

u' = 3x² + 6x

v' is the differential of v so:

v' = 1/x

Now we simply put the values into our product rule equation and put into our product rule equation:

u v' + u v'

(x³ + 3x²)(1/x) + (3x² + 6x)(ln(x))

Expanding the first set of brackets gives:

x² + 3x + (3x² + 6x)(ln(x))

Putting this into the equation stated in the question gives:

x² + 3x + (3x² + 6x)(ln(x)) = 2x² + 5x

Rearrange and simplify as follows:

(3x² + 6x)(ln(x)) = x² + 2x

ln(x) = (x² + 2x) / (3x² + 6x)

Now if we factorise the bottom of the fraction:

ln(x) = (x² + 2x) / (2(x² + 3x))

And cancelling the x² + 2x terms:

ln(x) = 1/3

Solving for x gives:

x = e^(1/3)

Integrate cos^2(x)

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