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Using integration by parts split it into v'=(1/x^3) and u=ln(x). v=-1/2x^2 and u'=1/x. Integral ln(x)/(x^3) = u*v - Integral u'*v = -ln(x)*1/2x^2 - 1/4x^2 + c
In order to integrate the expression we must first rewrite it in terms of Partial Fractions i.e. A/x and B/(2-x), so that when multiplied together we have a fraction with same denominator as the expressio...
Label the equations (1): y=x/2 + 2 and (2): 2y+3x=12. Firstly, multiply equation (1)by 2, this will give you (1a): 2y=x+4, then, substitute (1a) into (2). This gives you (x+4)+3x=12. Therefore, 4x+4=12. 4...
First, we need to factorise the equation on the left hand side, this can be done by finding two numbers that add together to make the 'b' coefficient (-9) and multiply to make the 'c' coefficient (20).
Due to notation difficulties, S = integral sign. This is a straightforward integration question, firstly we expand the brackets:
f'(x)=3x^2 - 3x f(x)=S(3x^2 - 3x)dx f(x)=3/3x^3 - 3/2x^2 + C f(x)=x^...
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