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∫ u(dv/dx) dx = uv − ∫ v(du /dx)dx is the Integration by Parts formula.
If you set u=lnx, differentiation (rememeber from tables) leads to du/dx= 1/x, and dv/dx=x and so v=x^2/2 (raise power by one then divide by that).
Plugging this into the equation, f(x)=(x^2/2)lnx- ∫(x^2/2)/x dx, just taking the RHS integral -> 1/2∫x dx = x^2/4 +C and so combining all of this f(x)=(x^2/2)lnx-x^2/4 +C.

Answered by Minty M.

Studies BioMedical Engineering at Imperial College London

This type of question appears over-complicated with limited options, but one must not fear! At first, the numerator seems to be of a higher degree than the denominator (x^4 compared to x^2), but from the first two terms of our numerator we can factorise out a common x^2 from (x^4-x^2) to give x^2(x^2-1) and thus factorise by the difference of two squares hence bringing our numerator to: x^2(x-1)(x+1) +2. By this, we can cancel of x^2(x-1) from both numerator and denominator leaving: x + 1 + 2/(x^2(x-1)) to be integrated.
Since we have three terms separated by addition, we can separate our integral in the following 3 separate integrals which we will add together at the end: (1) integrate x ; (2) integrate 1 ; (3) integrate 2/(x^2(x-1)). Beginning with (3), once again we can turn a complicated format into something simpler, this time we will use partial fractions. To use partial fractions, you must first distinguish how many terms lie in the denominator. From our denominator, x^2(x-1) we may initially believe that we only have 2 terms, but upon reflection of the rules for partial fractions, we have 3. This is since, under patricidal fractions we must have only linear factors in the denominator, thus from x^2 we derive two denominators: x and x^2. So in total we have the three denominators of x, x^2 and (x-1). Using partials fractions we can form the equality A/x + B/x^2 + C/(x-1) = 2/(x^2(x-1)). Simplifying both sides gives Ax^2 -Ax +Bx -B +Cx^2 = 2. By solving simultaneous equations, we find A=B= -2 and C=2. We can now create a simpler integral for 2/(x^2(x-1)), that being -2/x -2/x^2 +2/(x-1). Once we include integral (1) and (2), we come to the final simplified set of terms before integration: -2/x -2/x^2 +2/(x-1) +x +1. Using the principals of the natural log (ln) where the integral of 1/x = ln(x), we can integrate: -2/x = -2ln(x), while the integral of 2/(x-1) = 2ln(x-1). The integral of -2/x^2 =2/x because the negative exponent cancels out the minus sign while the integral of x+1 = (x^2)/ 2 +x. Finally bringing it all together our final answer is: -2lnx + 2/x +2ln(x-1) +(x^2)/2 +x + C. The C stands for constant which is to always be included in the answered derived from integration.

Answered by Nicholas H.

Studies BSc HONS Actuarial Science at LSE

Using the product rule, f=uv, df = (vu'-uv')/v^2. we first set u = 3x^2 and v = sin(2x). u' = 6x, v'=2cos(2x) Therefore, vu' = 6x*sin(2x). uv' = 6x^2*cos(2x), v^2 = 4cos^2(2x) Therefore the differential is [6x*sin(2x) - 6x^2*cos(2x)]/[4*cos^2(2x)] We can factor out 6x from the top and divide by the 4 on the bottom to give 3x(sin(2x)-x*cos(2x))/(2*cos^2(2x))

Look at each of the x variables to determine what happens to each term.
3x^2 has a power of 2 on the variable, therefore, the 2 is multiplied by the coefficient on the x. You must also subtract 1 from the value of 2 on the indices. This leaves you with 6x^1 which can be written as 6x.
The same must be done for 5x, however as the 3 has no x variable, it is not present in dy/dx

Answered by Rajan S.

Studies Economics and Mathematics at Bristol

Integrate by parts.
First rewrite the integral in the form u*dv/dx, which is (1)ln(x).
Then integrate (1)ln(x) wrt dx by assigning u=ln(x) du/dx=1/x and dv/dx=1 v=x.
We can determine the integral of ln(x), using the following formula for integration by parts:
integral of u*dv/dx wrt x = (u*v) − (integral of v*du/dx wrt x ).

To differentiate composite functions, (a function within a function) like in this case we need to use the chain rule.
We can see that F(x)=f(g(x)) where we let f(x)= (x^2+1)^2 and g(x)= x^2+1. To use the chain rule we need to find f'(x) and g'(x). The derivative of both functions.
To find the derivative of f(x) we let u= x^2+1 so f(u) becomes: u^2 so when we differentiate it we get f'(u)= 2u. Now we substitute u back in. So f'(x)=2(x^2+1). And g'(x)= 2x, we just differentiate it normally.
Now we put it all together, the chain rule says: F'(x)= f'(g(x))g'(x) so F'(x)= 2(x^2+1)2x = __4x(x^2+1)__

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